SAT Subject Test in Physics - SAT Subject Test in Physics

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Question

A disc of copper is dropped into a glass of water. If the copper was at $85^\circ C$ and the water was at $70^\circ C$ , what is the new temperature of the mixture?

$c_{Cu}=0.092\frac{cal}{g^oC}$

$c_{water}=1.00\frac{cal}{g^oC}$

Answer

The relationship between mass and temperature, when two masses are mixed together, is:

$m_1c_1\Delta T=-m_2c_2\Delta T$

Using the given values for the mass and specific heat of each compound, we can solve for the final temperature.

$(2g)(0.092\frac{cal}{g^\circ C})(T_f-85^\circ C)=-(12g)(1\frac{cal}{g^\circ C})(T_f-70^{\circ} C)$

We need to work to isolate the final temperature.

$(0.184\frac{cal}{^\circ C})(T_f-85^\circ C)=-(12\frac{cal}{^\circ C})(T_f-70^{\circ} C)$

Distribute into the parenthesis using multiplication.

$(0.184\frac{cal}{^\circ C})T_f-15.64cal=(-12\frac{cal}{^\circ C})T_f+840cal$

Combine like terms.

$(12.184\frac{cal}{^oC})T_f=855.64cal\textup{}$

$T_f=\frac{855.64cal}{12.184\frac{cal}{^oC}}$

$T_f=70.23^\circ C$

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Question

A capacitor with capacitance is constructed by putting a thin piece of cardboard between two copper plates, then each plate is connected to a battery. If the copper plates are cut in half then what is the effect on the capacitance of the circuit?

Answer

Capacitance of a circuit is defined by the equation:

$C=\varepsilon\frac{A}{d}$

Where is the capacitance, $\varepsilon$ is a constant of nature, is the area of the capacitor, and is the distance between the two plates

Since the metal plates are cut in half, the area is halved. We can substitute $\frac{1}{2}A$ in for

The result is that the capacitance is half the original quantity.

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Question

A book falls off the top of a bookshelf. What is its velocity right before it hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Answer

The relationship between velocity and energy is:

$KE=\frac{1}{2}mv^2$

We know the mass, but we need to find the total kinetic energy.

Remember the law of conservation of energy: the total energy at the beginning equals the total energy at the end. In this case, we have only potential energy at the beginning and only kinetic energy at the end. (The initial velocity is zero, and the final height is zero).

If we can find the potential energy, we can find the kinetic energy. The formula for potential energy is .

Using our given values for the mass, height, and gravity, we can solve using multiplication. Note that the height becomes negative because the book is traveling in the downward direction.

$PE=2kg-9.8\frac{m}{s^2}-2.3m$

The kinetic energy will also equal , due to conservation of energy.

Using this value and our given mass, we can calculate the velocity from our original kinetic energy equation.

$KE=\frac{1}{2}mv^2$

$45.08J=\frac{1}{2}2kgv^2$

$45.08\frac{m^2}{s^2}=v^2$

$\sqrt{45.08\frac{m^2}{s^2}}=\sqrt{v^2}$

Since we are taking the square root, our answer can be either negative or positive. The final velocity of the book will be in the downward direction; thus, our final velocity should be negative.

$-6.71\frac{m}{s}=v$

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Question

If the distance between two charged particles is doubled, the strength of the electric force between them will __________.

Answer

Coulomb's law gives the relationship between the force of an electric field and the distance between two charges:

$F_e=k\frac{q_1q_2}{r^2}$

The strength of the force will be inversely proportional to the square of the distance between the charges.

When the distance between the charges is doubled, the total force will be divided by four (quartered).

$F_e=k\frac{q_1q_2}{(2r)^2}=k\frac{q_1q_2}{4r^2}=\frac{1}{4}(k\frac{q_1q_2}{r^2})$

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Question

A model rocket, launched vertically, travels upwards and falls to the ground. At what point during flight is the rocket's acceleration greatest?

Answer

Newton's second law tells us that force and acceleration are directly related; if there is an acceleration, then there is also a force. This principle can help conceptualize this question.

While the rocket is in the air, there is only one force acting on it: the force of gravity. We can thus conclude that the acceleration of the rocket is directly related to this force. Since the force on the rocket (the force of gravity) is constant, its acceleration is also constant.

Any object that is in projectile or free-fall motion will experience a constant acceleration due to gravity.

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Question

A train blows its whistle as it approaches a station at $17\frac{m}{s}$ . If the sound emitted by the whistle has a frequency of , what is the frequency of the sound heard by the people at the station?

$v_s=340\frac{m}{s}$

Answer

Wave frequency is distorted when the observer or source is in motion based on the Doppler effect. The equation for the Doppler effect is:

$f=f_o\frac{v_s\pm v_{obs}}{v_s\pm v_{source}}$

Terms in the numerator are summed if the observer moves toward the source. Terms in the denominator are summed if the source is moving away from the observer.

In this question, the train (source) is moving toward the observer. This means that the denominator terms will be subtracted. The velocity of the observer is zero, giving us our final equation:

$f=f_o\frac{v_s}{v_s-v_{source}}$

Use the given values for the emitted frequency, velocity of the train, and speed of sound to solve this equation.

$f=(570Hz)\frac{340\frac{m}{s}}{340\frac{m}{s}-17\frac{m}{s}}$

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Question

A dog is stationary in someone's front yard and barks at a frequency of . Which of the following could explain why an observer hears a frequency of ?

Answer

The doppler effect states that the observed frequency depends on the frequency produced, and the relative velocities of the observer and sound source:

$f_{o}= f\frac{v\pm v_{o}}{v\pm v_{s}}$

To increase the observed frequency the sound source and the observer must be moving towards each other. Since the dog is stationary, therefore the observer must be moving towards the dog.

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Question

On which of the following does the amount of work required to move a charge in an electric field depend?

Answer

Work done by electric field is defined:

$W=-\Delta PE$

Notice that the only variable in the equation is the potential, so this is the only quantity on which work depends.

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Question

Of the following, which type of waves has the longest wavelength?

Answer

Wavelength and frequency have an inverse relationship; as one value increases, the other value decreases. In contrast, frequency and wave energy have a direct relationship; as one term increases, the other increases as well.

In the electromagnetic spectrum all wave types have the same velocity, but some have greater energy (and frequency) than others. Gamma rays have the highest energy, while radio waves have the lowest. In order from lowest energy to highest, the types of electromagnetic waves are radio waves, microwaves, infrared light, visible light, ultraviolet light, X-rays, and gamma rays.

Of the given answer options, microwaves have the lowest energy. This means they will also have the lowest frequency, and largest wavelength.

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Question

A crate slides across a floor for before coming to rest from its original position.

What is the force due to friction?

Answer

Since there is only one force acting upon the object, the force due to friction, we can find its value using the equation . The problem gives us the mass of the crate, but we have to solve for the acceleration.

Start by finding the initial velocity. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

$\Delta x=\frac{(v_f+v_i)}{2}\Delta t$

Plug in our given values and solve.

$10m=\frac{(0\frac{m}{s}+v_i)}{2}5s$

$\frac{10m}{5s}=\frac{v_i}{2}$

$2\frac{m}{s}=\frac{v_i}{2}$

$4\frac{m}{s}=v_i$

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

$\Delta x=v_it+\frac{1}{2}at^2$

Plug in the given values to solve for acceleration.

$10m=(4\frac{m}{s}) (5s)+\frac{1}{2}a(5s)^2$

$10m=(20m)+\frac{1}{2}a(25s^2)$

$-10m=\frac{1}{2}a(25s^2)$

$\frac{-20m}{25s^2}=a$

$-0.8\frac{m}{s^2}=a$

Now that we have the acceleration and the mass, we can solve for the force of friction.

$F=(2.9kg)(-0.8\frac{m}{s^2})$

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Question

If the distance between two objects is reduced by two-thirds, how will the gravitational force between the objects be affected?

Answer

According to Newton's law of universal gravitation, the gravitational force between two objects is inversely proportional to the square of the distance between them.

$F_G=G\frac{m_1m_2}{d^2}$

If the distance is reduced by two-thirds, then the final distance is equal to one-third the original distance.

$d_2=\frac{1}{3}d_1$

Using this term in the equation will show that the force increases by a factor of nine.

$F_G=G\frac{m_1m_2}{(\frac{1}{3}d_1)^2}=G\frac{m_1m_2}{\frac{1}{9}d_1^2}=9(G\frac{m_1m_2}{d_1^2})$

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Question

An egg falls from a nest in a tree that is tall. A girl, away, runs to catch the egg. If she catches it right at the moment before it hits the ground, how fast does she need to run?

$\small g=-9.8\frac{m}{s^2}$

Answer

The important thing to recognize here is that the amount of time the egg is falling will be equal to the amount of time the girl is running.

Our first step will be to find the time that the egg is in the air.

We know it starts from rest above the ground, and we know the gravitational acceleration. Its total displacement will be , since it falls in the downward direction. We can use the appropriate motion equation to solve for the time:

$\Delta y =v_it+\frac{1}{2}at^2$

Use the given values in the formula to solve for the time.

$-2.8m=(0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-2.8m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-2.8m=(-4.9\frac{m}{s^2})t^2$

$\frac{-2.8m}{-4.9\frac{m}{s^2}}=t^2$

Now that we have the time, we can use it to find the speed of the girl. Her speed will be determined by the distance she travels in this amount of time.

$v=\frac{d}{t}$

Use our values for her distance and the time to solve for her velocity.

$v=\frac{13m}{0.755s}$

$v=17.22\frac{m}{s}$

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Question

A ball is dropped from the roof a building that is tall. How long will it be before the ball hits the ground?

Answer

We can solve this problem using the kinematics equations. Note that the initial velocity will be zero, since the ball is dropped, and the acceleration will be equal to the acceleration of gravity.

$d=v_ot+\frac{1}{2}at^2$

This can be simplified, since the initial velocity is zero.

$d=\frac{1}{2}at^2$

Use the given value for the distance (the height of the building) and the acceleration of gravity to solve for the time.

$80m=\frac{1}{2}(9.8\frac{m}{s^2})t^2$

$\frac{80m}{\frac{1}{2}(9.8\frac{m}{s^2})}=16.3s^2=t^2$

$t=\sqrt{16.3s^2}=4.0s$

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Question

If the block of mass slides up a frictionless incline plane and is pulled by mass which is falling then what is the acceleration of the block on the ramp if and and the angle of incline is 30 degrees?

Answer

Begin by making a free body diagram for each block:

Use the diagram to write an equation for net force on each block:

$F_{net} =F_{t}-mg \sin (30)$

$F_{net} = Mg -F_{t}$

Since for the block on the ramp and for falling block we can substitute into these equations:

$ma =F_{t}-mg \sin (30)$

$Ma = Mg -F_{t}$

Then add the equations to get:

$Ma+ma = Mg-mg\sin (30)$

Next rearrange to isolate :

$a = \frac{M-msin(30)}{M+m}$

Substitute the given values from the question and solve:

$a = \frac{40-20sin(30)}{40+20}$

$a=\frac{1}{2}g$

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Question

A cart has a linear momentum with a magnitude of $20\frac{kg\cdot m}{s}$ . What is the cart's kinetic energy?

Answer

Linear momentum is calculated as the product of mass and velocity:

We are given the mass of the cart and its momentum, allowing us to solve for its velocity.

$20\frac{kg\cdot m}{s}=(4kg)v$

$v=\frac{20\frac{kg\cdot m}{s}}{4kg}=5\frac{m}{s}$

Now that we know the velocity of the cart, we need to use the equation for kinetic energy:

$KE=\frac{1}{2}mv^2$

Use the value of the velocity and the given mass of the cart to solve.

$KE=\frac{1}{2}(4kg)(5\frac{m}{s})^2$

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Question

A ball hits a brick wall with a velocity of $30\frac{m}{s}$ and bounces back at the same speed. If the ball is in contact with the wall for , what is the value of the force exerted by the wall on the ball?

Answer

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

$mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})(\frac{s}{s})$

$\frac{kgm}{s^2}s=Ns=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\Delta p=F\Delta t$ can also be thought of as $(p_{final}-p_{initial})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{final}-mv_{initial})=F\Delta t$

Even though the ball is bouncing back at the same "speed" its velocity will now be negative, as it is moving in the opposite direction. Using this understanding we can solve for the force in our equation.

$(0.25kg* -30\frac{m}{s})-(0.25kg* 30\frac{m}{s})=F(0.1s)$

$(-7.5\frac{kg\cdot m}{s})-(7.5\frac{kg\cdot m}{s})=F(0.1s)$

$(-15\frac{kg\cdot m}{s})=F(0.1s)$

$\frac{(-15\frac{kg\cdot m}{s})}{0.1s}=F$

Our answer is negative because the force is moving the ball in the OPPOSITE direction from the way it was originally heading.

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Question

Near the surface of the earth, a projectile is fired from a canon at an angle of $\theta$ degrees above the horizon and an initial velocity of meters per second. Which of the following expressions gives the time it takes the projectile to reach its maximum height?

Answer

At the maximum height of projectile motion, $v_{y}=0$ and because this takes place near the surface of the earth we know which we can plug into the equation:

$v_{y} = v_{0} + at$

$0 = v_{0} + gt$

We can then rearrange for

$t = \frac{v_{0}}{g}$

Next substitute this value into the equation $v_{y} = v \sin \theta$ to get the correct answer

$t = v \sin \frac{\theta }{g}$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

How long is the rock in the air?

Answer

We first need to find the vertical component of the velocity.

$v_{i}\sin(\Theta)=v_{iy}$

We can plug in the given values for the angle and the initial velocity to find the vertical component.

$30\frac{m}{s} \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

Now we need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

$0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1$

$\frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the vertical CHANGE. Since the rock started at the top of a building, if it rose an extra , then at its highest point it is above the ground.

This means that our $\Delta y$ will be as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

$\Delta y =v_it+\frac{1}{2}at^2$

$-25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=(-4.9\frac{m}{s^2})t^2$

$\frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{5.12s^2}=\sqrt{t^2}$

Add together the time for upward travel and downward travel to find the total flight time.

$t_{total}=t_1+t_2$

$t_{total}=1.76s+2.26s$

$t_{total}=4.02s$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

What is the horizontal distance that the rock travels?

Answer

We first need to find the horizontal component of the initial velocity.

$v_{i}* \cos(\Theta)=v_{x}$

We can plug in the given values for the angle and initial velocity and solve.

$30\frac{m}{s}* \cos(35^{\circ})=v_{x}$

$30\frac{m}{s}* (0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

The only force acting on the rock during flight is gravity; there are no forces in the horizontal direction, meaning that the horizontal velocity will remain constant. We can set up a simple equation to find the relationship between distance traveled and the velocity.

$\Delta x =v_xt$

We know , but now we need to find the time the rock is in the air.

We need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

$0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1$

$\frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the vertical CHANGE. Since the rock started at the top of a building, if it rose an extra , then at its highest point it is above the ground.

This means that our $\Delta y$ will be as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

$\Delta y =v_it+\frac{1}{2}at^2$

$-25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=(-4.9\frac{m}{s^2})t^2$

$\frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{5.12s^2}=\sqrt{t^2}$

Add together the time for upward travel and downward travel to find the total flight time.

$t_{total}=t_1+t_2$

$t_{total}=1.76s+2.26s$

$t_{total}=4.02s$

Now that we've finally found our time, we can plug that back into the equation from the beginning of the problem, along with our horizontal velocity, to solve for the final distance.

$\Delta x =v_xt$

$\Delta x =24.57\frac{m}{s}* 4.02s$

$\Delta x=98.77m$

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

At what angle to the horizontal will the rock impact the ground?

Answer

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i} \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

The horizontal velocity will not change during flight because there are no forces in the horizontal direction. The vertical velocity, however, will be affected. We need to solve for the final vertical velocity, then combine the vertical and horizontal vectors to find the total final velocity.

We know that the rock is going to travel a net distance of , as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity.

$v_f^2=v_1^2+2a\Delta y$

$v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}$

$v_f^2=492.18\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}$

$v_f=22.19\frac{m}{s}$

Because the rock is traveling downward, our velocity will be negative: $v_f=-22.19\frac{m}{s}$ .

Now that we know our final velocities in both the horizontal and vertical directions, we can find the angle created between the two trajectories. The horizontal and vertical velocities can be compared using trigonometry.

$\tan(\Theta)=\frac{v_y}{v_x}$ ,

$\tan^{-1}(\frac{v_y}{v_x})=\Theta$

Plug in our values and solve for the angle.

$\tan^{-1}(\frac{22.19\frac{m}{s}}{24.57\frac{m}{s}})=\Theta$

$\tan^{-1}(0.90)=\Theta$

$41.99^{\circ}=\Theta$

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