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In the diagram above, a mass m starting at point A is projected with the same initial horizontal velocity v0 along each of the three tracks shown here (with negligible friction) sufficient in each case to allow the mass to reach the end of the track at point B. (Path 1 is directed up, path 2 is directed horizontal, and path 3 is directed down.) The masses remain in contact with the tracks throughout their motions. The displacement A to B is the same in each case, and the total path length of path 1 and 3 are equal. If t1,t2, and t3 are the total travel times between A and B for paths 1, 2, and 3, respectively, what is the relation among these times?
A. t3 < t2 < t1
B. t2 < t3 < t1
C. t2 < t1 = t3
D. t2 = t3 < t1
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A: All the masses begin with the same speed v0. For Path 1, the object must climb up which would decrease its speed, then come back down to reach its same initial speed. During this time, the speed is always below the initial speed. For Path 2, the object maintains the same initial speed the whole time. For Path 3, the object speeds up as it goes down the path and then slows down back to its initial speed when it climbs. During this time, the speed is always above the initial speed. If we compare the average speed, v3> v2> v1, hence t3< t2< t1.
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Physics/76
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Physics
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The graph above shows the velocities of two objects undergoing a head-on collision. Given that Object 1 has 4 times the mass of Object 2, which type of collision is it?
A. Perfectly elastic
B. Perfectly inelastic
C. Inelastic
D. Cannot be determined
|
A: First, perfectly inelastic can be immediately eliminated because the objects have different velocities after the collision. Next, recall that a collision is perfectly elastic if kinetic energy is conserved.
\[ K_0 = K_f \]
\[ \frac{1}{2} m_{1,0} v_{1,0}^2 + \frac{1}{2} m_{2,0} v_{2,0}^2 = \frac{1}{2} m_{1,f} v_{1,f}^2 + \frac{1}{2} m_{2,f} v_{2,f}^2 \]
\[ \frac{1}{2} m_1 (6 \text{ m/s})^2 + \frac{1}{2} (4 m_1)(1 \text{ m/s})^2 = 20 m_1 \text{ J} = 20 m_1 \text{ J} \]
Because kinetic energy was conserved, the collision must be perfectly elastic.
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Physics/77
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Physics
| null |
Determine the total power dissipated through the circuit shown above in terms of v, R1, R2, and R3.
A. $ \frac{V^2}{R_1 + R_2 + R_3} $
B. $ \frac{R_1 + R_2 + R_3}{V^2} $
C. $ \frac{R_1 (R_2 + R_3)}{V^2 (R_1 + R_2 + R_3)} $
D. $ \frac{V^2 (R_1 + R_2 + R_3)}{R_1 (R_2 + R_3)} $
|
D: First, we will need to replace all the resistors with an equivalent resistor.
\[ R_2 \text{ and } R_3 \text{ are in series, so } R_{2 \text{ and } 3} = R_2 + R_3. \]
\[ R_1 \text{ and } R_{2 \text{ and } 3} \text{ are in parallel, so } \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_{2} + R_{3}} \text{ or } R_{\text{eq}} = \frac{R_1 (R_2 + R_3)}{R_1 + R_2 + R_3}. \]
Second, we need to calculate the current coming out of the battery,
\[ I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{V}{\frac{R_1 (R_2 + R_3)}{R_1 + R_2 + R_3}}. \]
Finally, we calculate the power,
\[ P = IV = \left( \frac{V^2 (R_1 + R_2 + R_3)}{R_1 (R_2 + R_3)} \right). \]
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Physics/78
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Physics
| null |
If v = 100 V, R1 = 50 Ω, R2 = 80 Ω and R3 = 120 Ω, determine the voltage across R3.
A. 100 V
B. 60 V
C. 40 V
D. 20 V
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B: Because R1 and R2 and R3 are in parallel, the voltage across each matches the battery of 100 V. Using what we solved from problem 39 and our knowledge that current splits off in parallel, to calculate the current going into R2 and R3 we use
I = V/R = V/R2 and 3 = V/(R2 + R3) = (100 V)/(80 Ω + 120 Ω) = (100 V)/(200 Ω) = 0.5 A
Now that we know the current going into this branch of the parallel circuit, we know that R2 and R3 are in series, and in series the current remains the same. So the current going across R3 is 0.5 A. Using this we can calculate the voltage going across R3,
V3 = I3R3 = (0.5 A)(120 Ω) = 60 V
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Physics/79
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Physics
| null |
Question below refers to the following information:A rigid rod of length L and mass M sits at rest on an air table with negligible friction. A small blob of putty with a mass of m moves to the right on the same table, as shown in overhead view in the figure. The putty hits and sticks to the rod, a distance of 2L/3 from the top end.How will the rod-putty system move after the collision?
A. The system will have no translational motion, but it will rotate about the rod’s center of mass.
B. The system will move to the right and rotate about the rod-putty system’s center of mass.
C. The system will move to the right and rotate about the rod’s center of mass.
D. The system will have no translational motion, but it will rotate about the rod-putty system’s center of mass.
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B: Conservation of linear momentum requires that the center of mass of the system continue to move to the right after the collision. The rotation will be about the combined rod-putty center of mass. To understand that, imagine if the putty were really heavy. Then after the collision, the rod would seem to rotate about the putty, because the center of mass of the rod-putty system would be essentially at the putty's location. In this case, you don't know whether the putty or the rod is more massive, but you do know that when the two objects stick together, they will rotate about wherever their combined center of mass is located.
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Physics/80
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Physics
| null |
Question below refers to the following information:A rigid rod of length L and mass M sits at rest on an air table with negligible friction. A small blob of putty with a mass of m moves to the right on the same table, as shown in overhead view in the figure. The putty hits and sticks to the rod, a distance of 2L/3 from the top end.What quantities, if any, must be conserved in this collision?
A. Linear momentum only
B. Neither linear nor angular momentum
C. Angular momentum only
D. Linear and angular momentum
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D: No unbalanced forces act here other than the putty on the rod and the rod on the putty. (The weight of these objects is canceled by the normal force.) Thus, linear momentum is conserved. No torques act on the rod-putty system except those due to each other; thus, angular momentum is conserved. It's essentially a fact of physics that in a collision between two objects, both linear and angular momentum must be conserved.
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Physics/81
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Physics
| null |
Bob and Tom hold a rod with a length of 8 m and weight of 500 N. Initially, Bob and Tom each hold the rod 2 m from the its ends, as shown in the figure. Next, Tom moves slowly toward the right edge of the rod, maintaining his hold. As Tom moves to the right, what happens to the torque about the rod’s midpoint exerted by each person?
A. Bob’s torque decreases, and Tom’s torque increases.
B. Bob’s torque increases, and Tom’s torque decreases.
C. Both Bob’s and Tom’s torque increases.
D. Both Bob’s and Tom’s torque decreases.
|
C: The net torque on the rod must be zero, because the rod doesn't rotate. Therefore, whatever happens to Bob's torque must also happen to Tom's torque-these torques must cancel each other out. That eliminates choices A and B. The forces provided by Bob and Tom must add up to 500 N, the weight of the rod. Try doing two quick calculations: In the original case, Bob and Tom must each bear 250 N of weight and are each 2 m from the midpoint, for 500 N·m of torque each. Now put Tom farther from the midpoint-say, 3 m away. For the torques to balance, Fbob(2 m) = Ftom(3 m). The only way to satisfy this equation and get both forces to add up to 500 N is to use 300 N for Fbob and 200 N for Ftom. Now, the torque provided by each is (300 N)(2 m) = 600 N·m.
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Physics/82
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Physics
| null |
An object of mass m hangs from two ropes at unequal angles, as shown in the figure. Which of the following makes correct comparisons between the horizontal and vertical components of the tension in each rope?
A. Horizontal tension is equal in both ropes, but vertical tension is greater in rope A.
B. Both horizontal and vertical tension are equal in both ropes
C. Horizontal tension is greater in rope B, but vertical tension is equal in both ropes.
D. Both horizontal and vertical tension are greater in rope B
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A: The object is in equilibrium, so left forces equal right forces. Thus, the horizontal tensions must be the same in each rope. Rope A pulls at a steeper angle than rope B, but with the same amount of horizontal force as rope B. To get to that steeper angle, the vertical component of the tension in rope A must be larger than in rope B.
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Physics/83
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Physics
| null |
Question below refers to the following information:Block B is at rest on a smooth tabletop. It is attached to a long spring, which in turn is anchored to the wall. Identical block A slides toward and collides with block
B. Consider two collisions, each of which occupies a duration of about 0.10 s:Collision I: Block A bounces back off of block B.Collision II: Block A sticks to block B.In which collision, if either, does block B move faster immediately after the collision?
A. In collision I, because block A experiences a larger change in momentum, and conservation of momentum requires that block B does as well.
B. In collision I, because block A experiences a larger change in kinetic energy, and conservation of energy requires that block B does as well.
C. In neither collision, because conservation of momentum requires that both blocks must have the same momentum as each other in each collision.
D. In neither collision, because conservation of momentum requires that both blocks must change their momentum by the same amount in each collision.
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A: There's no indication that energy must be conserved in collision 1. However, momentum is always conserved in a collision. When block A bounces, its momentum has to change to zero and then change even more to go back the other way. Since block A changes momentum by more in collision I, block B must as well because conservation means that any momentum change by block A must be picked up by block B. Choices C and D are wrong because, among other things, they use conservation of momentum to draw conclusions about two separate collisions; momentum conservation means that total momentum remains the same before and after a single collision, not in all possible collisions.
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Physics/84
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Physics
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If two people pull with a force of 1000 N each on opposite ends of a rope and neither person moves, what is the magnitude of tension in the rope?
A. 0 N
B. 500 N
C. 1000 N
D. 2000 N
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C: Just focus on one end of the rope. Nothing is moving, so the net force must be 0. If a person pulls with a force of 1000 N and nothing moves, then the resulting tension must also be 1000 N. Additionally, tension is a constant magnitude throughout a string. The direction of the tension can change (as in the case of a pulley system), but the magnitude will remain the same.
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Physics/85
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Physics
| null |
Two identical blocks are stacked on top of each other and placed on a table. To overcome the force of static friction, a force of 10 N is required. If the blocks were placed side by side and pushed as shown in the figure above, how much force would be required to move them?
A. $\frac{10 \sqrt{2}}{2 \mathrm{~N}}$
B. 10 N
C. $10 \sqrt{2} \mathrm{~N}$
D. 20 N
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B: The force of static friction will be $F_f=\mu F_N=\mu(m g)$. Changing the arrangement of the blocks does not change any of these three quantities, so the force will remain the same. Thus, $10 \mathrm{~N}$ will again be required to move them.
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Physics/86
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Physics
| null |
A block of known mass M is connected to a horizontal spring that is sliding along a flat, frictionless surface. There is an additional block of known mass m resting on top of the first block. Which of the following quantities would NOT be needed to determine whether the top block will slide off the bottom block?
A. The maximum coefficient of static friction between the blocks
B. The amplitude of the system's motion
C. The spring constant
D. The average speed of the blocks
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D: It's easiest to consider both blocks as one system, calculate the maximum acceleration of the pair of blocks caused by the spring (which occurs at maximum displacement from equilibrium), and determine whether friction between the two block is strong enough to give the top block the necessary acceleration to keep up with the bottom block. Applying Newton's Second Law to the system of blocks, $F_{\text {spring }}=$ (combined mass) ${ }^* a$ $\rightarrow k A=(M+m) a$. In order for the top block to accelerate at this rate, static friction must be strong enough that $F_{\mathrm{sf}}=m a \rightarrow \mu \mathrm{s} F \mathrm{~N}=m a$. You don't need to do any math to see that $\mu \mathrm{s}$ (coefficient of static friction), $A$ (amplitude), and $k$ (spring constant) appear, but average speed doesn't.
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Physics/87
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Physics
| null |
Question below refers to the below circuit diagram.If switch S1 is connected to point A but switch S2 is left unconnected, what is the current through the 1-Ω resistor?
A. 0 A
B. 2 A
C. 3 A
D. 72/11 A
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B: With $S_1$ closed but $S_2$ open, only the left-hand loop is connected. The resistors are in series, so the equivalent resistance is $R_{\text {eq }}=2 \Omega+1 \Omega+3 \Omega=6 \Omega$. The current through the circuits including the $1 \Omega$ resistor is $I_{\text {left }}=$ $\frac{V_{\text {left }}}{R_{\text {eq }}}=\frac{12 \mathrm{~V}}{6 \Omega}=2 \mathrm{~A}$.
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Physics/88
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Physics
| null |
Question below refers to the below circuit diagram.If switch S2 is connected to point B but switch S1 is left unconnected, what is the current through the 1-Ω resistor?
A. 0 A
B. 2 A
C. 3 A
D. 288/19 A
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C: With $S_1$ open but $S_2$ closed, only the right-hand loop is connected. Again, the resistors are in series so the equivalent resistances is $R_{\text {eq }}=4 \Omega+1 \Omega+3 \Omega=8 \Omega$. The current through the circuit including the $1 \Omega$, I_{\text{right}} = \frac{V_{\text{right}}}{R_{\text{eq}}} = \frac{24\ V}{8\ \Omega} = 3\ A.
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Physics/89
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Physics
| null |
Question below refers to the below circuit diagram.If both switches S1 and S2 are left in the unconnected positions, what is the current through the 1-Ω resistor?
A. 0 A
B. 1 A
C. 5 A
D. 36 A
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A: If neither switch is connected, then no loop can be made, which means no current will flow.
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Physics/90
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Physics
| null |
In real projectile motion, an object experiences three forces: gravity, drag, and lift. These are depicted in the picture above. Given this information, how does lift affect the speed of a projectile?
A. Increases
B. Decreases
C. Varies
D. It has no effect.
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D: Because the lift force is perpendicular to the velocity of the object, lift can have no effect on the magnitude of velocity. The magnitude of velocity is also known as speed, so (D) is correct.
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Physics/91
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Physics
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The picture above shows a tube open at one end. A standing wave is represented by the dotted curves. If this wave has frequency f, what is the frequency of the next harmonic that can be formed in this tube?
A. 1/2 f
B. 2f
C. 3f
D. The above wave shows the highest possible harmonic frequency for this system.
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C: Remember that in a system with one closed end and one open end, there will be a node at one end and an antinode at the other. From the drawing, you can see that the tube is the length of 1/4 wavelength of this standing wave. That means that this wave is the fundamental standing wave, also called the first harmonic. As you'll learn on this page, the next harmonic is the third harmonic, and it has 3 times the frequency of the first harmonic.
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Physics/92
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Physics
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Which of the following correctly describes an electron moving from point A to point B in the situation above? Assume the two regions of charge are identical in magnitude and only different in sign.
A. The electron moves with increasing speed and gains electric potential energy.
B. The electron moves with increasing speed and loses electric potential energy.
C. The electron moves with decreasing speed and gains electric potential energy.
D. The electron moves with decreasing speed and loses electric potential energy.
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B: An electron is repelled from negative charge and attracted to positive charge, so there are two electric forces on it, both pointing to the right. Thus, as it moves from A to B, the electron's speed increases. This eliminates (C) and (D). As for potential energy: one way to think about it is to realize that an increase in speed implies an increase in kinetic energy. Where does this energy come from? It comes from the potential energy of the system, which decreases by an equal amount.
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Physics/93
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Physics
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What is the equivalent resistance of the electric circuit in the picture above?
A. 0.46 Ω
B. 1.2 Ω
C. 4.5 Ω
D. 14.5 Ω
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D: Remember that parallel resistors require you to use the reciprocal sum, while series resistors simply add. For example, the first set of parallel resistors would be combined by doing $\frac{1}{R}=\frac{1}{1.5}+\frac{1}{3}=1$, which means $R=1 \Omega$. For the second set of parallel resistors, you would get $\frac{1}{R}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}$, which means $R=1.5 \Omega$. Adding those two values to the lone series resistor gives $R_{\text {Total }}=2+1+1.5=4.5 \Omega$.
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Physics/94
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Physics
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Escape velocity is defined as the minimum speed at which an object must be launched to "break free" from a massive body's gravitational pull. Which of the following principles could be used to derive this speed for a given planet?
A. Conservation of Linear Momentum
B. Newton's Third Law
C. Conservation of Angular Momentum
D. Conservation of Energy
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D: If an object leaves a planet at exactly escape velocity, its speed will approach zero as its distance from the planet approaches infinity. Zero speed implies zero KE, and infinite separation implies zero Ug if our reference separation is infinity. Thus, the total mechanical energy of an object traveling at escape velocity is 0 J. That's why conservation of energy is the relevant concept.
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Physics/95
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Physics
| null |
An ambulance is driving toward you. As it approaches, which of the following correctly describes the changes in the sound of the siren's pitch and intensity?
A. Increasing pitch, increasing intensity
B. Increasing pitch, decreasing intensity
C. Decreasing pitch, increasing intensity
D. Decreasing pitch, decreasing intensity
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A: The intensity of a wave is proportional to $1 / r^2$, so it will increase as the ambulance approaches. This eliminates (B) and (D). The change in pitch will be determined by the Doppler effect. Pitch is simply another term for frequency, and the Doppler effect says that if the source of a wave and the object detecting the wave are growing closer, then the frequency will increase. Thus, $(\mathrm{A})$ is correct.
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Physics/96
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Physics
| null |
The graph above depicts a medium's instantaneous displacement from equilibrium, caused by a wave, as a function of distance from the source. If the wave has a speed of 600 m/s, which of the following is the best approximation of the wave's frequency?
A. 50 Hz
B. 100 Hz
C. 200 Hz
D. Cannot be determined without additional information
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B: The distance from one crest to another is the wavelength. In this case, that distance is 6 m. Furthermore, you know that v = fλ for any wave. Solving for f gives f = v/λ. Plugging in the known values gives f = (600)/(6) = 100 Hz.
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Physics/97
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Physics
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Both of the above strings have their ends locked in place. The two strings have the same linear density, but the first string, S1, is twice as long as the second string, S2. If sound waves are going to be sent through both, what is the correct ratio of the fundamental frequency of S1 to the fundamental frequency of S2?
A. 2:1
B. $\sqrt{2}: 1$
C. $1: \sqrt{2}$
D. 1:2
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D: A wave passing along a string with both ends held in place will have a fundamental frequency of f = v/(2l), where v is the speed of the wave and l is the length of the string. Thus, fundamental frequency is inversely proportional to the length of the string, so the first string, which is twice as long as the second, will have a fundamental frequency that is half of the second string's. Expressed as a ratio, that is 1:2.
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Physics/98
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Physics
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A pendulum with a ball of mass m hanging from a string of length l is set in motion on Earth, and the system is found to have a frequency of f. If the length of the string were doubled, the hanging mass tripled, and the system moved to the moon, what would be the new frequency?NOTE: Acceleration due to gravity of the Moon is approximately of Earth's.
A. $\frac{1}{12} f$
B. $\sqrt{\frac{1}{12}} f$
C. $\sqrt{12} f$
D. 12f
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B: The frequency of a pendulum is \( f = \frac{1}{2\pi} \sqrt{\frac{g}{l}} \). Therefore, the change in mass would have no effect on the system, and the others would change the equation to \( f = \frac{1}{2\pi} \sqrt{\frac{g}{2l}} \). Thus, the original frequency \( f \) will become \( \frac{1}{\sqrt{12}} f \).
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Physics/99
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Physics
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A block of mass M is at rest on a table. It is connected by a string and pulley system to a block of mass m hanging off the edge of the table. Assume the hanging mass is heavy enough to make the resting block move. Knowing the acceleration of the system and the mass of each block is sufficient to calculate all of the following quantities EXCEPT which one?
A. Net force on each block
B. Tension in the string
C. Coefficient of kinetic friction between the table and the block of mass M
D. The speed of the block of mass M when it reaches the edge of the table
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D: The net force on each block can be found by using Newton's Second Law, \( F_{\text{Net}} = ma \). The tension in the string can be found by focusing on the hanging mass. You know the net force on it will be \( F_{\text{Net}} = ma = F_g - T = (mg) - T \), which means \( T = (mg) - (ma) \). Finally, the coefficient of kinetic friction can be found by looking at the top block. For that block, the net force in the horizontal direction will be the same as the overall net force since the two vertical forces (normal and gravity) will cancel out. Thus, you get \( F_{\text{Net}} = ma = T - F_f = T - \mu F_N = T - \mu mg \). The only value in the answers you cannot calculate is the speed of the block when it reaches the edge. In order to compute this value, you would need to know how far from the edge the block is when it begins moving.
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Physics/100
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Physics
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A block of mass m is connected by a string which runs over a frictionless pulley to a heavier block of mass M. The smaller block rests on an inclined plane of angle θ, and the larger block hangs over the edge, as shown above. In order to prevent the blocks from moving, the coefficient of static friction must be
A. $\frac{mgsin\theta}{Mg-mg\cos\theta}$
B. $\frac{Mg-mgsin\theta}{Mg\cos\theta}$
C. $\frac{Mg-mgsin\theta}{mg\cos\theta}$
D. $\frac{Mg-mg\cos\theta}{mg\sin\theta}$
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C: If nothing is moving, then you know that the net force will be 0. Looking first at the forces perpendicular to the plane, you get \( F_N = F_g \cos\theta = mg\cos\theta \). Next, using Newton's Second Law and defining 'up the ramp' as positive, you can say \( F_{\text{Net}} = T - F_g \sin\theta - F_f = 0 \). Solving for tension and plugging in all the variables gives you \( T = mgsin\theta + \mu mg\cos\theta \).
Next, looking at the hanging block, you can again use Newton's Second Law to determine that \( F_{\text{Net}} = F_g - T = Mg - (mgsin\theta + \mu mg\cos\theta) = 0 \). Thus, solving for \(\mu\) gives you \(\mu = \frac{Mg - mgsin\theta}{mg\cos\theta}\).
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Physics/101
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Physics
| null |
A comet orbits the Sun in a periodic elliptical orbit as shown below:A comet orbits the Sun in a periodic elliptical orbit as shown below:
A comet orbits the Sun in a periodic elliptical orbit as shown below:What happens to the gravitational potential energy and the kinetic energy, respectively, of the comet-Sun system as the comet moves from point A to point B?
A comet orbits the Sun in a periodic elliptical orbit as shown below:Ug K
A. Increases Increases
B. Decreases Decreases
C. Increases Decreases
D. Decreases Increases
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(D) As the comet moves toward the Sun, the gravitational potential energy, https://img.apstudy.net/ap/physics-1/a500/Ans_466.jpg decreases (becomes more negative) as the radial distance r, from the Sun decreases. Since there is no net force on the comet-Sun system, its mechanical energy must remain constant. Thus, with the gravitational potential energy decreasing, the kinetic energy must increase.
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Physics/102
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Physics
| null |
A comet orbits the Sun in a periodic elliptical orbit as shown below:A comet orbits the Sun in a periodic elliptical orbit as shown below:
A comet orbits the Sun in a periodic elliptical orbit as shown below:As the comet moves from point A to point B, what happens to the comet's angular momentum about the Sun?
A. Increases linearly
B. Increases proportionally to the square of the distance from the Sun
C. Decreases
D. Stays constant
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(D) The gravitational force from the Sun is the only force acting on the comet. Since this force is pointed directly at the Sun, it will not have a lever arm to apply a net torque to the comet system about the Sun. With zero net torque on the comet, its angular momentum must stay constant for angular momentum to be conserved.
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Physics/103
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Physics
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Four identical rods shown above experience the forces as shown. Rank the magnitude of the torques about the pivot point on the left end of the rod.
A. III > I = IV > II
B. II > IV > III > I
C. I = III = IV > II
D. III > II > I > IV
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(C) Torque is calculated by https://img.apstudy.net/ap/physics-1/a500/Ans_461_1.jpg Assuming the rod length L, the individual torques are calculated as follows: https://img.apstudy.net/ap/physics-1/a500/Ans_461_2.jpg The rank is as follows: https://img.apstudy.net/ap/physics-1/a500/Ans_461_3.jpg
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Physics/104
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Physics
| null |
The diagram below shows a top-view of a rod that is free to rotate about its center and is initially rotating with a positive counter-clockwise angular velocity. Two forces are applied to the rod that are steady in magnitude and will continue to act perpendicular to the rod, even after it rotates.The diagram below shows a top-view of a rod that is free to rotate about its center and is initially rotating with a positive counter-clockwise angular velocity. Two forces are applied to the rod that are steady in magnitude and will continue to act perpendicular to the rod, even after it rotates.
The diagram below shows a top-view of a rod that is free to rotate about its center and is initially rotating with a positive counter-clockwise angular velocity. Two forces are applied to the rod that are steady in magnitude and will continue to act perpendicular to the rod, even after it rotates.What value of the unknown downward force will result in a constant angular velocity of the rod?
A. F
B. F/2
C. 2F
D. 4F
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(C) In order for angular velocity to be constant, there must be zero net torque on the rod: https://img.apstudy.net/ap/physics-1/a500/Ans_452.jpg
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Physics/105
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Physics
| null |
The figure above shows four identically shaped disks, each with a fixed axis of rotation at their centers. Disk IV has twice the rotational inertia of disks I, II, and III. The disks are subjected to a variety of forces as shown. Rank the magnitude of the angular acceleration of the disks.
A. III > II = IV > I
B. I = II = III = IV
C. II = IV > I = III
D. II > I = III = IV
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(D) Angular acceleration is the ratio of net torque to rotational inertia: https://img.apstudy.net/ap/physics-1/a500/Ans_446_1.jpg . Assume the forces act at a distance R from the axis of rotation. https://img.apstudy.net/ap/physics-1/a500/Ans_446_2.jpg https://img.apstudy.net/ap/physics-1/a500/Ans_446_3.jpg
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Physics/106
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Physics
| null |
Questions below are based on the following figure of a mass-spring system. Assume the mass is pulled back to position +A and released, and it slides back and forth without friction.At what position does the mass have the greatest acceleration?
A. -A
B. -A/2
C. 0
D. +A/2
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(A) At position -A and +A, the compression/extension of the spring is the greatest distance from equilibrium, giving the largest magnitude of force. This force at position -A is to the right (the positive direction) and yields the largest positive net force. Thus, acceleration, the ratio of net force to mass, is at a maximum at this position.
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Physics/107
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Questions below are based on the following figure of a mass-spring system. Assume the mass is pulled back to position +A and released, and it slides back and forth without friction.When the mass reaches position 0, what can be said about its speed?
A. It is at its minimum.
B. It is at its maximum.
C. It is zero.
D. It is decreasing.
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(B) As the mass passes through equilibrium, it has accelerated to its maximum speed.
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Physics/108
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In the physics lab, a hooked mass is hung from a string as shown above. A photogate is set up at the bottom of the swing that provides a signal to the computer when and infrared beam (represented by the dashed line on the diagram) is blocked or unblocked by the hooked mass. How can the photogate signal be used to measure the oscillation period of the swing of the hooked mass?
A. Divide the diameter of the hooked mass by the elapsed time between when the beam is blocked and subsequently unblocked.
B. Measure the elapsed time between when the beam is blocked and subsequently unblocked.
C. Measure the elapsed time between when the beam is blocked and subsequently blocked again.
D. Double the elapsed time between when the beam is blocked and subsequently blocked again.
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(D) The period of a pendulum is defined as the time for one complete back and forth swing. When the hooked mass gets to the bottom of the swing, the clock will start when the beam gets blocked. If the clock is turned off the next time the bob blocks the beam, it has only swung through half the swing. Thus, the period may be calculated by doubling the elapsed time between when the beam is blocked and subsequently blocked again.
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Physics/109
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A spring-mass system hangs on a ring stand that rests on top of a cabinet as shown in the diagram above. The cabinet houses the air compressor for a physics lab. The compressor shakes the ring stand 0.2 cm up and down at a rate of 6 vibrations per second. Which of the following is a true statement?
A. The maximum vibration amplitude of the spring-mass system is 0.2 cm.
B. The spring-mass system will only vibrate for very high values of the spring constant.
C. The spring-mass system may vibrate with an amplitude considerably greater than 0.2 cm.
D. The spring-mass system will vibrate wildly if it has a natural frequency of 12 vibrations per second.
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(C) Resonance is a dramatic growth in vibration amplitude that occurs when a system is forced to vibrate at its natural frequency. The natural frequency of a spring-mass system depends on its mass and the spring constant. If the natural frequency of the spring mass system is 6 Hz, then it will resonate with amplitudes considerably greater than 0.2 cm.
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Physics/110
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The graph shows the displacement versus time for an object. Which equation best describes its displacement in meters?
A. Îx = 20 cos(0.5t)
B. Îx = 10 cos(2t)
C. Îx = 10 cos(Ït)
D. Îx = 20 sin(Ït)
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(C) The general equation for this oscillation is Îx = A cos(2Ïf t), where A is the amplitude and f is the frequency of vibration. According to the graph, a complete cycle occurs every 2 s which is a frequency of 0.5 cycles per second. The graph also shows an amplitude of vibration of 10 m. This gives the following equation: Îx = 10 cos(Ït).
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Physics/111
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An isolated, stationary sphere of mass 4m explodes into three fragments as shown in the figure above. The fragment with mass 2m moves vertically upward with a speed V and the fragment with mass m moves to the left with speed 2V. What is the magnitude and direction of the momentum of the third fragment?
Magnitude of momentum Direction
A. 2mV â
B. â2(mV) â
C. 2â2(mV) â
D. 0 None
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(C) Because there is no net force on the sphere during the explosion, the momentum must be conserved for the system. Since the system has no momentum before the explosion, the total momentum after the explosion must be zero. The following figure may be used to visualize the postexplosion momenta: https://img.apstudy.net/ap/physics-1/a500/Answer_360.jpg The arrow up and to the left represents the combined momenta of the first two fragments. To obtain zero net momentum, the third fragment must have an equal momentum down and to the right. Its value may be obtained from the Pythagorean theorem as follows: https://img.apstudy.net/ap/physics-1/a500/Ans_028.jpg
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Physics/112
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As shown in the diagram above, a block of mass m and speed vo has an elastic collision with a second block of unknown mass that is originally at rest. The first block bounces back in the opposite direction at half its original speed. Is it possible to find the unknown mass and the final velocity of the second block in terms of the given quantities?
A. No. Momentum conservation may be applied to this collision, but there are too many unknowns to find a solution.
B. Yes. Momentum is conserved as well as mechanical energy. Both principles may be used together to solve this problem.
C. Yes. Momentum conservation alone is sufficient to determine the unknown mass and velocity.
D. Yes. Energy conservation alone is sufficient to determine the unknown mass and velocity.
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(B) Because this is an elastic collision, the kinetic energy in the two-body system before the collision must be equal to the kinetic energy in the two-body system after the collision. This provides the first equation with two unknowns. Momentum conservation may also be used to develop a second equation with the same two unknowns. Two equations are sufficient to solve a problem with two unknowns.
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Physics/113
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A 4-kg lab cart is moving to the right at a velocity of +5 m/s and has a head-on collision with a 2-kg lab cart moving at an unknown initial velocity. After a perfectly inelastic collision, the system moves to the right with a velocity of +2 m/s. Determine the velocity of the 2-kg cart before the collision.
A. +4 m/s
B. -4 m/s
C. +8 m/s
D. -8 m/s
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(B) Momentum is conserved as follows: https://img.apstudy.net/ap/physics-1/a500/An0_265.jpg
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Physics/114
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A 50.0-g ball moving to the left strikes a wall and bounces back to the right. Slow-motion video analysis produces the following horizontal position versus time graph.A 50.0-g ball moving to the left strikes a wall and bounces back to the right. Slow-motion video analysis produces the following horizontal position versus time graph.
A 50.0-g ball moving to the left strikes a wall and bounces back to the right. Slow-motion video analysis produces the following horizontal position versus time graph.The video camera used in the experiment has a frame rate of 30 pictures per second. In the video, the ball makes contact with the floor for three frames. What is the magnitude and direction of the average force that the ball exerts on the wall?
A. 7.2 N to the right
B. 4.5 N to the right
C. 4.5 N to the left
D. 7.2 N to the left
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(C) With 30 frames per second, a picture is taken each thirtieth of a second. Three frames will be 3/30 of a second, or 0.1 second. To find the average force, solve the momentum-impulse theorem for force as follows: https://img.apstudy.net/ap/physics-1/a500/Ans_338.jpg . This is the force of the floor on the wall, which is in the positive direction (to the right). According to Newton's third law, the force of the ball on the wall is equal in magnitude and opposite in direction, or -4.5 N (to the left).
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Physics/115
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A 50.0-g ball moving to the left strikes a wall and bounces back to the right. Slow-motion video analysis produces the following horizontal position versus time graph.A 50.0-g ball moving to the left strikes a wall and bounces back to the right. Slow-motion video analysis produces the following horizontal position versus time graph.
A 50.0-g ball moving to the left strikes a wall and bounces back to the right. Slow-motion video analysis produces the following horizontal position versus time graph.Determine the momentum change of the bouncing ball.
A. +0.05 kg à m/s
B. +0.45 kg à m/s
C. -0.45 kg à m/s
D. -0.05 kg à m/s
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(B) First the initial and final velocities are determined from the slope of the graph: https://img.apstudy.net/ap/physics-1/a500/A_09.jpg Next, calculate the momentum change: https://img.apstudy.net/ap/physics-1/a500/Ans_337.jpg
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Physics/116
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The graph above of momentum versus time depicts the motion of a box being pushed across a horizontal floor with negligible friction. Which of the following statements describes the force upon the box?
A. The force on the box is constant.
B. The force on the box is decreasing.
C. The force on the box is increasing.
D. More information is needed to describe the force on the box.
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(C) The slope of a momentum-time graph is force. Another way of looking at this is with the momentum-impulse theorem: The change in momentum divided by the time interval is the average force exerted on an object: https://img.apstudy.net/ap/physics-1/a500/An_04.jpg . Notice that the momentum is changing more and more rapidly as time elapses (i.e., the slope gets greater), which means the force is increasing.
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Physics/117
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A 500-kg spaceship in deep space is holding a 50-kg space probe, and the pair is initially drifting to the right at 20 m/s as shown in the first diagram above. If the spaceship fires the probe to the left at 40 m/s, what is the new speed of the spaceship?
A. 9 m/s
B. 13 m/s
C. 18 m/s
D. 26 m/s
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(D) Because there is no net force on the spacecraft/probe system, momentum is conserved: https://img.apstudy.net/ap/physics-1/a500/An000_7.jpg
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Physics/118
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Physics
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A 1-kg block of dry ice moving at 6 m/s collides and sticks to a 2-kg block of dry ice initially at rest. The combination slides off a 20-m tall cliff as shown in the diagram above. What is the horizontal distance from the base of the cliff to the landing location of the block combination?
A. 1 m
B. 2 m
C. 4 m
D. 5 m
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(B) Since there is no net force on the system during the collision, momentum conservation may be used to find the postcollision speed: https://img.apstudy.net/ap/physics-1/a500/BKN_17.jpg Next, the vertical free-fall time is found for the projectile: https://img.apstudy.net/ap/physics-1/a500/A_324.jpg During the 1-second free-fall time, the projectile moves at a constant horizontal speed with no horizontal acceleration: https://img.apstudy.net/ap/physics-1/a500/A_324A.jpg
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Physics/119
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An 1,800-kg truck moving north at 15 m/s has a head-on collision with a 900-kg car moving south at 22 m/s. During the collision, compare the magnitudes of the following quantities for the car versus the truck during the collision.
An 1,800-kg truck moving north at 15 m/s has a head-on collision with a 900-kg car moving south at 22 m/s. During the collision, compare the magnitudes of the following quantities for the car versus the truck during the collision.Force Acceleration Momentum change
A. Same Same Same
B. Same Different Same
C. Different Same Different
D. Different Different Different
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(B) Newton's third law states that the magnitudes of the forces are the same during a collision, regardless of the mass or speeds of the objects involved. With the same forces and different masses, the net force to mass ratio will be different, so the accelerations will be different according to Newton's second law. With the same forces and the same time of contact, the magnitude of the impulse will be the same for both vehicles, and impulse is equal to the momentum change according to the impulse-momentum theorem.
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Physics/120
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Physics
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The graph above depicts how the momentum of a box pushed across the floor changes with time. Which of the following statements describes the force on the box?
A. The force on the box is a constant 0.5 N.
B. The force on the box is a constant 1 N.
C. The force on the box is a constant 2 N.
D. The force on the box is 2 N and increasing.
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(C) Newton defined momentum as the time rate of change of force, so the slope of a momentum-time graph is force. Another way of looking at this is using the momentum-impulse theorem: The change in momentum divided by the time interval is the average force exerted on an object: https://img.apstudy.net/ap/physics-1/a500/A_306.jpg
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Physics/121
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Physics
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A person pulls a 25-kg block of dry ice (initially at rest) with a changing force as shown in the diagram and graph above. The block moves a distance of 6.0 m along a frictionless table in 3.2 s of time. Determine the power output of the person.
A. 56 W
B. 75 W
C. 90 W
D. 180 W
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(A) Power is the time rate at which the person works on the system. The person does work to give the block kinetic energy, and this work can be found by finding the average force times the displacement (or the area bounded by the force-displacement graph). https://img.apstudy.net/ap/physics-1/a500/f0265-02.jpg
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Physics/122
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Physics
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In its initial state, the 75-kg wrecking ball shown in the diagram above is released from rest at a distance of 3.5 m off the ground. After swinging back and forth multiple times, it's captured in its final state swinging up at a speed of 5.0 m/s at the instant it is 2.0 m off the ground. Determine the total work done by dissipative forces (such as air drag and pivot friction) on the pendulum system between its initial and final states.
A. 0 J
B. -190 J
C. -560 J
D. -1,700 J
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(B) The total work done on the system by dissipative forces is the change in mechanical energy of the system. https://img.apstudy.net/ap/physics-1/a500/f0264-01.jpg
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Physics/123
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Physics
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A tall sailing ship (m = 1.43 à 104 kg) is initially at rest. A steady wind blows at the 60° angle shown in the diagram and moves the ship forward a distance of 850 m. The ship's final speed is 9.0 m/s. Assuming that water resistance is negligible, what is the force of the wind on the boat?
A. 200 N
B. 390 N
C. 680 N
D. 790 N
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(D) The work done by the wind equals the change in kinetic energy of the boat: https://img.apstudy.net/ap/physics-1/a500/BK_17.jpg The total force of the wind, F, is found as follows: https://img.apstudy.net/ap/physics-1/a500/BK_18.jpg
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Physics/124
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Physics
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A 5-kg box slides 10 m diagonally down a frictionless ramp inclined at 45°. At the bottom of the ramp, it slides on a rough horizontal concrete floor with a coefficient of friction of 0.6.A 5-kg box slides 10 m diagonally down a frictionless ramp inclined at 45°. At the bottom of the ramp, it slides on a rough horizontal concrete floor with a coefficient of friction of 0.6.
A 5-kg box slides 10 m diagonally down a frictionless ramp inclined at 45°. At the bottom of the ramp, it slides on a rough horizontal concrete floor with a coefficient of friction of 0.6.How far does the box travel on the concrete before coming to a complete stop?
A. 6 m
B. 8 m
C. 10 m
D. 12 m
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(D) The force of friction works against the motion of the box to bring it to a stop. First calculate the force of friction, and then use the work-energy theorem to find the distance the box travels: https://img.apstudy.net/ap/physics-1/a500/A_242.jpg https://img.apstudy.net/ap/physics-1/a500/A_242a.jpg
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Physics/125
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Physics
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A 5-kg box slides 10 m diagonally down a frictionless ramp inclined at 45°. At the bottom of the ramp, it slides on a rough horizontal concrete floor with a coefficient of friction of 0.6.A 5-kg box slides 10 m diagonally down a frictionless ramp inclined at 45°. At the bottom of the ramp, it slides on a rough horizontal concrete floor with a coefficient of friction of 0.6.
A 5-kg box slides 10 m diagonally down a frictionless ramp inclined at 45°. At the bottom of the ramp, it slides on a rough horizontal concrete floor with a coefficient of friction of 0.6.What is the speed of the box as it reaches the bottom of the ramp?
A. 6 m/s
B. 10 m/s
C. 12 m/s
D. 14 m/s
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(C) Use conservation of energy to find the velocity of the box: https://img.apstudy.net/ap/physics-1/a500/A_223.jpg https://img.apstudy.net/ap/physics-1/a500/A_241A.jpg
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Physics/126
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Physics
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A spacecraft in deep space is isolated from its surroundings and is initially moving with the velocity v shown in the diagram above. It has thrusters on a swivel that fire such that the spacecraft experiences three possible forces that act at fixed angles relative to the velocity vector of the spacecraft. With each of the forces separately with a steady magnitude, what initially happens to the kinetic energy K, of the spacecraft?
F1 F2 F3
A. K decreases K constant K increases
B. K increases K increases K increases
C. K decreases K constant K decreases
D. K increases K decreases K decreases
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(A) When F1 is applied, a vertical component of the force acts opposite the velocity vector and slows the spacecraft down, but the horizontal component only changes its direction because it's perpendicular to the velocity. As the spacecraft's speed slows, K decreases. When F2 is applied, it only changes the spacecraft's direction because it remains perpendicular to the velocity vector, thus K will not change. When F3 is applied, the vertical component of force acts as a net force in the same direction of the velocity vector and speeds the spacecraft up, but the horizontal component only changes its direction because it's perpendicular to the velocity. As the spacecraft speeds up, K increases.
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Physics/127
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Physics
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An experiment was conducted with a hoop spring attached to a force probe, both of which were mounted to a low friction dynamics cart. The compression of a hoop spring may be modeled as a Hookean spring. The system was placed on a horizontal track with a motion encoder to measure the displacement of the cart. The following graph shows the experimental force vs. displacement for a black hoop (steeper line) and a white hoop, each tested separately. The white hoop/probe/cart mass is 0.500 kg and the black hoop/probe/cart mass is 1.000 kg.An experiment was conducted with a hoop spring attached to a force probe, both of which were mounted to a low friction dynamics cart. The compression of a hoop spring may be modeled as a Hookean spring. The system was placed on a horizontal track with a motion encoder to measure the displacement of the cart. The following graph shows the experimental force vs. displacement for a black hoop (steeper line) and a white hoop, each tested separately. The white hoop/probe/cart mass is 0.500 kg and the black hoop/probe/cart mass is 1.000 kg.
An experiment was conducted with a hoop spring attached to a force probe, both of which were mounted to a low friction dynamics cart. The compression of a hoop spring may be modeled as a Hookean spring. The system was placed on a horizontal track with a motion encoder to measure the displacement of the cart. The following graph shows the experimental force vs. displacement for a black hoop (steeper line) and a white hoop, each tested separately. The white hoop/probe/cart mass is 0.500 kg and the black hoop/probe/cart mass is 1.000 kg.How far must the white cart's hoop be compressed against the wall for it to leave at the same speed as the black-hoop cart moved in the previous question?
A. 5.0 cm
B. 5.7 cm
C. 6.2 cm
D. 8.1 cm
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(B) The final speed of the black-hoop cart was found in the previous solution: https://img.apstudy.net/ap/physics-1/a500/A_239.jpg To have the same speed as the black-hoop cart, the white-hoop cart needs a kinetic energy of: https://img.apstudy.net/ap/physics-1/a500/A_239A.jpg
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Physics/128
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Physics
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An experiment was conducted with a hoop spring attached to a force probe, both of which were mounted to a low friction dynamics cart. The compression of a hoop spring may be modeled as a Hookean spring. The system was placed on a horizontal track with a motion encoder to measure the displacement of the cart. The following graph shows the experimental force vs. displacement for a black hoop (steeper line) and a white hoop, each tested separately. The white hoop/probe/cart mass is 0.500 kg and the black hoop/probe/cart mass is 1.000 kg.An experiment was conducted with a hoop spring attached to a force probe, both of which were mounted to a low friction dynamics cart. The compression of a hoop spring may be modeled as a Hookean spring. The system was placed on a horizontal track with a motion encoder to measure the displacement of the cart. The following graph shows the experimental force vs. displacement for a black hoop (steeper line) and a white hoop, each tested separately. The white hoop/probe/cart mass is 0.500 kg and the black hoop/probe/cart mass is 1.000 kg.
An experiment was conducted with a hoop spring attached to a force probe, both of which were mounted to a low friction dynamics cart. The compression of a hoop spring may be modeled as a Hookean spring. The system was placed on a horizontal track with a motion encoder to measure the displacement of the cart. The following graph shows the experimental force vs. displacement for a black hoop (steeper line) and a white hoop, each tested separately. The white hoop/probe/cart mass is 0.500 kg and the black hoop/probe/cart mass is 1.000 kg.If the cart with the black hoop is pushed against a wall and compresses the spring 5.0 cm, how fast will it be released?
A. 0.22 m/s
B. 0.44 m/s
C. 0.53 m/s
D. 0.55 m/s
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(B) The elastic energy stored in the spring transfers to kinetic energy. https://img.apstudy.net/ap/physics-1/a500/BK_15.jpg
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Physics/129
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The four blocks shown in the figure above are released from the same height. Blocks B, C, and D are each released from rest, and block A is initially moving horizontally with a speed v. Blocks A and C each have a mass of 1 kg and blocks B and D are each 2 kg. Assuming air drag and incline friction are negligible, rank the speed of each block as it reaches the ground.
A. vA = vB > vC >vD
B. vA > vB = vC = vD
C. vA = vB = vC = vD
D. vB = vD > vA > vC
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(B) Because all four objects start at the same height relative to the ground, they all begin with the same gravitational energy. With negligible surface friction and air drag, all of that gravitational energy transfers to kinetic energy as they arrive at the ground level. Since objects B, C, and D all start at rest, the analysis looks like this: https://img.apstudy.net/ap/physics-1/a500/A_236.jpg Therefore, objects B, C, and D all arrive at the ground with the same speed (albeit at different times). In contrast, object A already has kinetic energy when it's released, so its analysis differs as follows: https://img.apstudy.net/ap/physics-1/a500/A_14.jpg Thus, object A arrives at a greater speed. The final ranking is as follows: https://img.apstudy.net/ap/physics-1/a500/A_236A.jpg
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A 3.0-kg object, initially moving to the right with a velocity of +4.0 m/s, experiences a positive net force that decreases linearly throughout the displacement as shown on the graph above. What is the kinetic energy of the object at the instant the net force is zero?
A. 24 J
B. 51 J
C. 99 J
D. 174 J
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(C) Work is defined as the area bounded by the net force-displacement graph: https://img.apstudy.net/ap/physics-1/a500/A_232.jpg
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Physics/131
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Physics
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A roller coaster is initially moving to the right as it approaches point A in the figure below. Assume that air drag and friction are negligible.A roller coaster is initially moving to the right as it approaches point A in the figure below. Assume that air drag and friction are negligible.
A roller coaster is initially moving to the right as it approaches point A in the figure below. Assume that air drag and friction are negligible.At which point is the total mechanical energy of the coaster-Earth system the greatest?
A. Location A
B. Location D
C. Location E
D. It is the same at all points.
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(D) The total mechanical energy is the sum of the kinetic energy and gravitational potential energy. It is the same at all points because there is no change in internal energy of the system (air drag and track friction are negligible).
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Physics/132
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Physics
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A roller coaster is initially moving to the right as it approaches point A in the figure below. Assume that air drag and friction are negligible.A roller coaster is initially moving to the right as it approaches point A in the figure below. Assume that air drag and friction are negligible.
A roller coaster is initially moving to the right as it approaches point A in the figure below. Assume that air drag and friction are negligible.At which location will the roller-coaster car move at the greatest speed?
A. Location E
B. Location B
C. Location C
D. Location D
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(A) The car will move the fastest at the point with the most kinetic energy, which is the lowest point, E, where most of the gravitational energy in the car-Earth system has transferred to the kinetic energy.
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Physics/133
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Physics
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A roller coaster is initially moving to the right as it approaches point A in the figure below. Assume that air drag and friction are negligible.A roller coaster is initially moving to the right as it approaches point A in the figure below. Assume that air drag and friction are negligible.
A roller coaster is initially moving to the right as it approaches point A in the figure below. Assume that air drag and friction are negligible.At which locations will a roller-coaster car have the same gravitational potential energy?
A. Locations A and E
B. Locations B and C
C. Locations C and D
D. Locations A and C
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(D) Points A and C are at the same height from the ground and thus have the same gravitational potential energy: Ug = mgÎy.
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Physics/134
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Physics
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The mass, turn radius, and speed of each car in the table below are shown relative to car A. Which of the following best ranks the centripetal force on the cars?The mass, turn radius, and speed of each car in the table below are shown relative to car A. Which of the following best ranks the centripetal force on the cars?
A. D > A = B > C
B. D > C = B > A
C. D = A = B > C
D. D = B > A > C
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(D) Centripetal force is the net force on the car toward the center of the circle and is calculated as follows: https://img.apstudy.net/ap/physics-1/a500/A_207.jpg Thus, D = B > A > C.
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Physics/135
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Physics
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The graph above depicts the tangential velocities of several circular space stations with different radii. All the stations are spinning. Which of the following statements is true?
A. The centripetal accelerations of the 3 shorter radii space stations are greater than 10 m/s2; those of the larger ones are less than 10 m/s2.
B. The centripetal accelerations of the 3 shorter radii space stations are greater than 5 m/s2; those of the larger ones are less than 5 m/s2.
C. The centripetal accelerations of all the stations are all nearly 5 m/s2.
D. The centripetal accelerations of all the stations are all nearly 10 m/s2.
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(D) Pick any two points on the graph and calculate the centripetal acceleration. Square the velocity, and divide it by the radius. You will find that all of them have approximately the same centripetal acceleration, i.e., 10 m/s2.
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Physics/136
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Physics
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The driver of a race car takes a turn on a horizontal track and is moving southeast at a particular instant as shown in the diagram above. If the car is gaining speed, which of the following is a possible direction of acceleration of the car at the instant shown in the diagram?
A. Southwest toward the center of the turn
B. Southeast in the direction of motion
C. South on the diagram
D. Northeast away from the center of the turn
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(C) Since the car is gaining speed, there is a southeast component of acceleration tangent to the curved path. But also, because the car is turning, there is a southwest component of acceleration toward the center of the turn (the centripetal acceleration). These two components sum as vectors to yield a resultant vector that is in the southerly direction. (Note: Depending on the magnitudes of the components, the resultant vector may have a slight easterly or westerly component, but the dominant direction will be south.)
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Physics/137
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Physics
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A car is moving over the top of a hill at a constant speed. Which of the following may be said about the vertical forces on the car in this scenario?
A. The normal force of the road is greater than the gravitational force from the Earth.
B. The normal force of the road is less than the gravitational force from the Earth.
C. The normal force of the road equals the gravitational force from the Earth.
D. The centrifugal force equals the gravitational force from the Earth.
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(B) Since the car is moving in a vertical circle, it must accelerate toward the center of the circle (centripetal acceleration). Since the center of the circle is downward relative to the car, the acceleration, and consequently the net force, is downward. This implies that the downward gravitational force from the Earth is greater than the upward normal force from the road.
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Physics/138
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Physics
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An empty 150-kg roller coaster cart is approaching a 6.0-meter-tall circular-shaped loop-the-loop as shown in the figure above. In order to complete the loop, determine the minimum speed of the cart when it is upside-down at the top of the loop.
A. 5.5 m/s
B. 7.7 m/s
C. 10 m/s
D. 12 m/s
|
(A) The minimum speed condition occurs when the cart loses contact with the track (normal force is zero) and the gravitational force solely provides the required centripetal force: https://img.apstudy.net/ap/physics-1/a500/Ans-N_054.jpg
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Physics/139
|
Physics
| null |
A rope accelerates a 20.0-kg block up a 30° ramp at a rate of 2.0 m/s/s. The kinetic friction coefficient between the block and the ramp is 0.18. Determine the value of the tension in the rope.
A. 71 N
B. 130 N
C. 170 N
D. 180 N
|
(C) The first free-body diagram below shows the forces on the object. The second diagram divides the weight vector into its components. Since the block is not accelerating perpendicular to the incline, the normal force will balance the cosine component of the weight. https://img.apstudy.net/ap/physics-1/a500/An0_060.jpg Newton's second law may be used to determine the tension as follows: https://img.apstudy.net/ap/physics-1/a500/f0221-01.jpg
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Physics/140
|
Physics
| null |
A force sensor interfaced with a computer pulls horizontally on a 2.0-kg box that is initially at rest on a horizontal surface. The box then breaks free and slides along the surface. The graph above shows how the measured force varies with time. Determine the value for the coefficient of static friction.
A. 0.3
B. 0.4
C. 0.5
D. 0.6
|
(B) The static friction coefficient may be found as the ratio of the maximum static friction force and the normal force. Since the box is on a horizontal surface and is not accelerating vertically, the normal force balances the weight, mg. During the region of the graph where the force increases from zero to the maximum, the box hasn't broken free; the force sensor measures the static friction because the box is in a state of constant velocity. According to the graph, the static friction force increases from zero to a maximum value of 8 N. The coefficient of static friction is calculated as follows: https://img.apstudy.net/ap/physics-1/a500/An0_059.jpg
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Physics/141
|
Physics
| null |
The sign in the elevator above is suspended by two cables. If the elevator is moving upward at a constant speed, select two of the following statements that are true.
A. The upward normal force on the sign is greater than the individual tensions in each rope.
B. The vertical component of the tension in rope #1 is greater than the downward weight of the sign.
C. The vertical component of the tension in rope #1 balances the downward weight of the sign.
D. The horizontal component of the tension in rope #1 balances the tension in rope #2.
|
(C, D) Since the sign is moving upward at a constant velocity (zero acceleration), there must be no net force on the sign; therefore, the forces on each axis must be balanced. There is no normal force on the sign.
|
Physics/142
|
Physics
| null |
A rope applies 35 N force as shown in the figure above. As a result, the box accelerates to the left along the surface. In addition to the gravitational force, which two of the following forces act on the box?
A. The downward force of the box on the table
B. The frictional force of the surface on the box
C. The upward force of the table on the box
D. The force of motion to the left on the box
|
(B, C) A free-body diagram would show an upward normal force, a downward gravitational force, a tension force up and to the left at a 30° angle, a frictional force to the right parallel to the surface, and possibly an air drag force to the right that is probably negligible. "Motion" is not a force, and the force of the block on the table is a valid force but does not act on the system of interest (which is the box).
|
Physics/143
|
Physics
| null |
The graph above depicts the velocity of a skydiver over time during a vertical fall through the air. Which of the following statements is true about the magnitude of the net force on the skydiver?
A. It increases until the acceleration reaches its maximum and the velocity becomes constant.
B. It decreases until the acceleration reaches zero and the velocity becomes constant.
C. It has a value of zero throughout the skydiver's free fall.
D. It increases until the acceleration reaches zero and the velocity becomes constant.
|
(B) The forces acting upon the skydiver are the downward gravitational force and upward air resistance force (drag). Initially, the net force is high (gravity > air resistance), the skydiver accelerates, and the velocity increases rapidly. Over time, the air resistance increases with speed, so the net force and acceleration decrease. At some point in time the skydiver reaches terminal velocity where there is no net force (gravity = air resistance), acceleration is zero, and the skydiver's velocity is constant.
|
Physics/144
|
Physics
| null |
An object's position with time is depicted in this graph. At which time range will there be nearly no net force acting on the object?
A. 0.5 to 1.0 s
B. 1.0 to 1.5 s
C. 1.8 to 2.2 s
D. 2.5 to 3.5 s
|
(C) A curve on the position versus clock reading graph indicates a changing slope, which is a changing velocity. The straight section between 1.8 and 2.2 s indicates a constant velocity. Newton's first law states that objects will maintain constant velocity as long as there's zero net force acting on it.
|
Physics/145
|
Physics
| null |
A box of unknown mass (m) slides down a plane inclined at an angle (θ) as shown in the diagram above. The plane has a coefficient of friction (μ). Which of the following expressions represents the rate of acceleration (a) of the box?
A. a = g(sinθ - μg cosθ)/m
B. a = g(cosθ - μg sinθ)/m
C. a = g(cosθ - μsinθ)
D. a = g(sinθ - μcosθ)
|
(D) Remember that the force of gravity (mg) must be resolved into x and y components, and that the force of friction (Ff) is related to the normal force (FN) by Ff = μFN. The box does not move above or into the plane, so the forces in the y direction must be balanced. Apply Newton's second law to both the x and y directions to come up with an expression for acceleration: https://img.apstudy.net/ap/physics-1/a500/Ans-N_039.jpg
|
Physics/146
|
Physics
| null |
The four blocks shown in the figure above are released from the same height above the ground. Blocks B, C, and D are each released from rest, and block A is initially moving horizontally with a speed v. The mass of blocks A and C is 1 kg and the mass of blocks B and D is 2 kg. Assuming air drag and surface friction are negligible, rank the time it takes each block to reach the ground.
A. tA = tC < tB = tD
B. tA < tB < tC < tD
C. tA = tB = tC = tD
D. tA = tB < tC < tD
|
(D) Blocks A and B are both free falling, and free-fall acceleration is independent of mass; therefore, they will hit the ground at the same time. The mass of blocks C and D will not affect their acceleration for the same reason that mass does not affect free-fall acceleration: the mass cancels out of the ratio of net force to mass in Newton's second law. Blocks C and D only have a component of gravitational force accelerating them along the incline, so they have less net force acting on them compared to blocks A and B and they must travel along the whole length of the incline, as compared with the direct path of block B. Thus blocks C and D will take more time to reach the ground. Because block C has a greater component of gravitational force than D and a shorter path to travel, its time of travel will be less than D.
|
Physics/147
|
Physics
| null |
An airplane uses a cable to tow a large banner advertisement as shown in the figure above. When the airplane cruises at a constant horizontal velocity through the air, which of the following correctly compares forces?
A. The tension force in the cable equals the gravitational force on the banner.
B. The forward force on the airplane equals the air drag force on the airplane and banner.
C. The y-component of cable tension equals the aerodynamic lift force on the airplane.
D. The x-component of cable tension is greater than the air drag force on the banner.
|
(B) A system that cruises at a constant velocity has no acceleration. As a consequence, Newton's second law predicts that the system will have no net force. Applying this to the horizontal forces on the airplane/banner system, the forward force on the airplane must equal the combined forces of air drag on the airplane and banner.
|
Physics/148
|
Physics
| null |
Problems below refer to the system of two boxes connected by a rope as shown below on the floor of an elevator. A force F is applied to the 6-kg box.Assuming negligible friction between the blocks and the elevator floor, determine the value of the tension in the rope between the boxes if F = 24 N.
A. 2 N
B. 6 N
C. 8 N
D. 20 N
|
(B) Use Newton's second law to calculate the acceleration of the entire 8-kg system: https://img.apstudy.net/ap/physics-1/a500/Ans-N_038.jpg Next, apply Newton's second law to the 6-kg subsystem: https://img.apstudy.net/ap/physics-1/a500/A_05.jpg Newton's second law may also be applied to the 2-kg subsystem, and the same value of tension will be found.
|
Physics/149
|
Physics
| null |
Problems below refer to the system of two boxes connected by a rope as shown below on the floor of an elevator. A force F is applied to the 6-kg box.Which of the following pair of forces ALWAYS have the same magnitude, regardless of the state of motion of the system?
A. The force of the rope pulling the 6-kg block to the right and the force of the 6-kg block pulling the rope to the left
B. The force F to the left and the force of the rope pulling the 6-kg box to the right
C. The normal force of the floor on the 6-kg block and the gravitational force on the 6-kg box
D. The force F to the left and the force of the rope on the 2-kg box to the left
|
(A) The force of the rope pulling the 6-kg block to the right and the force of the 6-kg block pulling the rope to the left form a Newton's third law pair. These forces will always be equal in magnitude and opposite in direction. The other force pairs may or may not be congruent, depending on the state of motion of the system and the presence or absence of friction. The normal force and the gravitational force on the 6-kg box will only be congruent if the vertical acceleration of the elevator system is zero.
|
Physics/150
|
Physics
| null |
Only two forces act on a 0.1-kg object. This graph depicts the magnitudes and directions of those two forces. What is the acceleration of the object at the 2-second clock reading?
A. 0 m/s2
B. 0.2 m/s2
C. 1 m/s2
D. 2 m/s2
|
(D) The object accelerates only when the forces acting upon it are unbalanced. The net force on the object at t = 2 s is 1.2 N - 1.0 N = 0.2 N. The acceleration is the net force divided by the mass, or (0.2 N/0.1 kg) = 2 m/s2.
|
Physics/151
|
Physics
| null |
The acceleration of a 5-kg object through 4 seconds of elapsed time is shown in the graph above. What is the net force on the object at the 1-second clock reading?
A. -10 N
B. -5 N
C. -2.5 N
D. 5 N
|
(A) From the graph, the acceleration at t = 1 s is - 2 m/s2. The object's mass is 5 kg. So, the net force can be calculated from Newton's second law: https://img.apstudy.net/ap/physics-1/a500/Ans-N_031.jpg
|
Physics/152
|
Physics
| null |
A boy is pushing a 50-kg crate across horizontal, frictionless rollers. The velocity is changing with time as shown in the graph above. What is the magnitude of the force that the boy applies to the crate?
A. 5 N
B. 10 N
C. 15 N
D. 25 N
|
(D) The acceleration is the slope of a velocity-time graph. In this case, the slope of the graph yields an acceleration of 0.5 m/s2. The mass of the crate is 50 kg. From Newton's second law, calculate the force applied: https://img.apstudy.net/ap/physics-1/a500/f0214-01.jpg
|
Physics/153
|
Physics
| null |
This position-time graph is typical of which type of motion?
A. Motion of an object with an increasing net force acting on it
B. Motion of an object with constant positive net force acting on it
C. Motion of an object with no net force acting on it
D. Motion of an object with negative net force acting on it
|
(C) The position in the graph is increasing linearly in the positive direction. It is consistent with motion with a constant positive velocity. The constant velocity means that acceleration is zero. According to Newton's second law, when acceleration is zero, there is no net force acting upon it.
|
Physics/154
|
Physics
| null |
Questions below refer to the 80-kg astronaut and her 40-kg daughter who are floating in deep space shown in the diagram below.Determine the magnitude of the mother's acceleration during her 160 N push on the daughter.
A. 0 m/s2
B. 0.5 m/s2
C. 1 m/s2
D. 2 m/s2
|
(D) In deep space, there is no gravitational force from a planet and the gravitational attraction between the bodies is negligible. Thus, the only force during the push is the 160-N force of the daughter pushing the mother to the left. Using Newton's second law, https://img.apstudy.net/ap/physics-1/a500/Ans-N_029.jpg to the left.
|
Physics/155
|
Physics
| null |
Questions below refer to the 80-kg astronaut and her 40-kg daughter who are floating in deep space shown in the diagram below.If the mother pushes her daughter to the right with a force of 160 N, with what magnitude of force does the daughter push her mother?
A. 0 N
B. 80 N
C. 160 N
D. 640 N
|
(C) This problem refers to a single pushing interaction between the mother and the daughter. The mother's push on the daughter forms a Newton's third law pair with the daughter's push on the mother, and thus forms an N3L pair that must be equal in magnitude and opposite in direction.
|
Physics/156
|
Physics
| null |
A box is pulled along a surface by a rope. The acceleration-time graph of an object's motion is shown in the figure above. At what time will the forces acting on the box be balanced?
A. 0 s
B. 1 s
C. 2 s
D. 3 s
|
(C) According to Newton's second law, the forces acting upon an object are balanced when the acceleration is zero. As shown in the graph, this occurs at 2 s.
|
Physics/157
|
Physics
| null |
The velocity versus time graph above compares the motion of object A with object B. Which of the following statements is true?
A. Both objects have the same acceleration at the 3-second clock reading.
B. Object B is changing directions near the 2-second clock reading.
C. Object A is slowing down while object B is speeding up throughout the 4-second trip.
D. The displacement of object B is greater than that of object A during the 4-second trip.
|
(B) Both objects demonstrate uniform acceleration throughout the trip as demonstrated by the constant slopes, but since the slopes are different in both sign and magnitude, the value of their accelerations are different. The sign (±) of the velocity determines the direction of motion. Object A is slowing down while moving in the positive direction because its positive velocity is decreasing. Object B starts with negative velocity (negative direction) and slows down for 2 seconds, then it speeds up in the positive direction.
|
Physics/158
|
Physics
| null |
The graph above shows position vs. time for an object. Which of the following best describes the motion of the object?
A. Gaining speed with a positive acceleration
B. Gaining speed with a negative acceleration
C. Losing speed with a positive acceleration
D. Losing speed with a negative acceleration
|
(B) Velocity is the slope of the position versus time graph. Because the graph starts out with a zero slope, the velocity is initially zero. As the slope gets more negative, the velocity gets more negative. Since speed is the absolute value of velocity, that means the speed is increasing. Acceleration is the slope of the velocity graph, and since the velocity is decreasing from zero to a negative value, the acceleration is negative.
|
Physics/159
|
Physics
| null |
The graph above shows the position of an object versus clock reading. Which of the following best describes the motion of the object during the 6 seconds of elapsed time?
A. The object speeds up for the first 2 seconds and slows down during the final 4 seconds.
B. The object travels a distance of 6 meters during the 6 seconds of elapsed time.
C. The object slows down and then speeds up with a total displacement of -6 meters.
D. The object's acceleration starts out negative and transitions to positive acceleration at the 2-second clock reading.
|
(C) Since the slope of the X versus t graph starts positive and goes to zero in the first 2 seconds, the object is slowing down. Then the slope increases negatively, indicating that the object is gaining speed in the negative direction. The total displacement is ÎX = Xf - Xi = 0 - 6m = -6m. (Note: Option D is incorrect because acceleration is defined as the slope of the velocity versus clock reading graph. This slope remains negative even after the 2-second clock reading, as the velocity transitions from zero to more negative values.)
|
Physics/160
|
Physics
| null |
The velocity-time graph of an object's motion is shown in this graph. At 10 s, what is the object's displacement relative to its position at t = 0?
A. 3 m
B. 6 m
C. 0 m
D. -6 m
|
(A) To find the displacement, integrate the area under the curve to the t axis. This involves adding the areas of two trapezoids, one in the 0-6 s time interval and the other in the 6-10 s time interval. https://img.apstudy.net/ap/physics-1/a500/Ans-N_022.jpg
|
Physics/161
|
Physics
| null |
Rank the average velocities of the car in regions A, B, C, and E of the graph above.
A. B > A = C > E
B. B = A > C = E
C. B > A = C = E
D. A > B > C = E
|
(A) The average velocity of region A is (0 + 30 m/s)/2 = +15 m/s. Region B is (30 m/s + 30 m/s)/2 = +30 m/s. Region C is (30 m/s + 0 m/s)/2 = +15 m/s. Region E is (0 m/s + -30 m/s)/2 = -15 m/s.
|
Physics/162
|
Physics
| null |
Which of the following are true about the motion of the carts in the graph above?
A. At the 7-second clock reading, cart 1 is moving with the same speed as cart 2.
B. Cart 1 moves at a constant speed in the negative direction.
C. Cart 1 moves in the negative direction over the entire 8-second trip.
D. At the 8-second clock reading, cart 1 was moving faster than cart 2.
|
(A, D) Option A is correct because speed is the absolute value of velocity, and both carts were moving at 4 m/s. Option D is also correct because, at the 8-second clock reading, Cart 1 was moving at 6 m/s and cart 2 was moving at 4 m/s. Option C is tempting but incorrect because it had positive velocity for the first half of the trip, indicating that it was moving in the positive direction.
|
Physics/163
|
Physics
| null |
Which of the following best describes the motion of the car in regions C, D, and E of the velocity versus time graph above?
A. It is slowing down until it reverses direction and speeds back up again.
B. It moves at a constant velocity in the negative direction.
C. It accelerates at a constant, nonzero acceleration except at point C when its acceleration is zero.
D. Its position changes at a constant rate.
|
(A) At the beginning of region C, the velocity gradually decreases to zero, indicating its speed is decreasing. At point D, the velocity has reached zero (resting instantaneously). At the beginning of region E, the negative velocity indicates that the direction is now negative, and the increasing magnitude of the velocity indicates it is speeding up.
|
Physics/164
|
Physics
| null |
An object's position versus time is depicted in the graph above. Based on the graph, at which time points will the object's velocity be closest to zero?
A. 0 s and 2 s
B. 0 s and 4 s
C. 1 s and 2 s
D. 1 s and 3 s
|
(D) When a position-time graph is horizontal, the velocity of the object is zero. The velocity at any instant in time is the slope of the line tangent to any point on the position-time graph. For this graph, that occurs at 1 s and 3 s.
|
Physics/165
|
Physics
| null |
The four different projectiles are fired from the surface of the Earth with negligible air drag. The following data table compares the four projectiles:Rank the vertical acceleration of the four projectiles.
A. a1 = a2 > a3 > a4
B. a4 < a2 < a3 < a4
C. a1 = a2 = a3 = a4
D. The masses must be known in order to rank the accelerations.
|
(C) The masses are all free falling since there is negligible air drag. Free-fall acceleration is independent of mass, so they have the same vertical acceleration.
|
Physics/166
|
Physics
| null |
The following diagrams show possible scenarios of a dynamics cart going through 2 photogates:Photogate #1 measures a 0.02-second elapsed time for the cart's vertical rod (diameter = 1.2 cm) to block the infrared beam and then unblock the same beam. Photogate #2 works in the same manner and measures a 0.06-second elapsed time for the rod to pass through the beam. Based on this data and the fact that the photogates are 20 cm apart, describe the motion of the cart and calculate the magnitude of its acceleration.
A. Speeding up with |a| = 0.5 m/s/s
B. Slowing down with |a| = 0.5 m/s/s
C. Speeding up with |a| = 0.8 m/s/s
D. Slowing down with |a| = 0.8 m/s/s
|
(D) https://img.apstudy.net/ap/physics-1/a500/Ans-N_006.jpg https://img.apstudy.net/ap/physics-1/a500/Asw_02.jpg
|
Physics/167
|
Physics
| null |
The position-time graphs of four different objects are shown in these graphs. If the positive direction is forward, then which object is moving backward at a constant velocity?
A. Object A
B. Object B
C. Object C
D. Object D
|
(D) If the positive direction is forward, then an object that is moving backward at a constant velocity would have a position vs. time graph that is linear with a negative slope.
|
Physics/168
|
Physics
| null |
This graph above depicts the motion of an object on a coordinate system for 10 s. During which time interval is the object moving at a constant positive velocity?
A. 0-2 s
B. 3-4 s
C. 4-6 s
D. 6-7 s
|
(A) The slope of a position-time graph is velocity. In the time interval from 0-2 seconds the slope is positive and constant, so the velocity is positive and constant in that same interval.
|
Physics/169
|
Physics
| null |
The position-time graph shown above is typical of which type of motion?
A. Motion with a constant negative velocity
B. Motion with zero velocity
C. Motion with a constant positive acceleration
D. Motion with zero acceleration
|
(D) The position in the graph is increasing linearly in the positive direction. This is consistent with motion with a constant positive velocity and, hence, zero acceleration.
|
Physics/170
|
Physics
| null |
For the diverging lens shown above, which principle rays are correctly drawn? Select two answers.
A. 1
B. 2
C. 3
D. 4
|
A, D For a diverging lens, the "focus" is on the far side of the lens. So (1) shows a light ray traveling into the focus and then bending parallel to the optic axis. (4) shows a light ray traveling parallel to the optic axis, and it leaves on a line connecting the other focal point.
|
Physics/171
|
Physics
| null |
The circuit shown above is set up. The switch is closed and a long time passes. What conditions on the two resistors result in the greatest amount of energy stored in the capacitor?
A. The energy stored in the capacitor will be greatest if R1 > R2.
B. The energy stored in the capacitor will be greatest if R1 = R2.
C. The energy stored in the capacitor will be greatest if R1 < R2.
D. The energy will be the same regardless of the resistor values.
|
D The energy stored in a capacitor depends on the voltage across the capacitor when it is fully charged. The capacitor will have the same voltage as the battery when fully charged regardless of the resistances of either resistor.
|
Physics/172
|
Physics
| null |
An experiment is conducted to determine the critical angle for light going from glass into air, as shown above. A linear plot is made with a vertical axis of sin (θ1) and a horizontal axis of sin (θ2). How is the critical angle determined from the graph?
A. The critical angle can be found from the slope of the line.
B. The critical angle can be found from the y-intercept of the line.
C. The critical angle can be found from the horizontal axis value which corresponds to the maximum vertical value.
D. The critical angle can be found from the vertical axis value which corresponds to the maximum horizontal value.
|
C The critical angle occurs at the angle for θ2 when the value of sin(θ1) = 1, which is the maximum value that sin (θ) can obtain.
|
Physics/173
|
Physics
| null |
Three cylinders of the same metal act as resistors arranged in series, as shown above. Which of the following correctly ranks the voltage drops across the three resistors?
A. V1 = V2 = V3
B. V3 > V1 > V2
C. V2 > V3 > V1
D. V1 > V3 > V2
|
C The resistance of a wire is directly proportional to the length of the wire and inversely proportional to the cross-sectional area. 1 must have the least resistance and 2 must have the most resistance. Because the resistors are in series, they have the same current through them. Ohm's Law states that the greatest voltage drop is across the greatest resistance value.
|
Physics/174
|
Physics
| null |
Charges are distributed as shown above. At point A is a charge of +3Q. At B is a charge of +1Q. At C is a charge of -1Q. What is the direction of the force on a proton located at point P?
A. Up and to the left
B. Down and to the left
C. Up and to the right
D. Down and to the right
|
C From the charge at point A, the field will be up and to the right. From B, the field will be up and to the left. The right component of A will be stronger than the leftward component of B. From the charge at C, the field will point down and to the right. The upward field lines from either A or B will be greater than the downward field lines from C.
|
Physics/175
|
Physics
| null |
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