SAT Subject Test in Physics

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SAT Subject Test in Physics › SAT Subject Test in Physics

Questions 1 - 10

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

Only the change in potential

Only the path traveled

Both the potential and the path traveled

Neither the potential nor the path traveled

Only the magnetic field

Explanation

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.

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

Only the change in potential

Only the path traveled

Both the potential and the path traveled

Neither the potential nor the path traveled

Only the magnetic field

Explanation

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.

If a circuit contains a battery and a $2\Omega$ resistor what is the current of the circuit?

$\frac{1}{5}A$

Explanation

Ohm's law states:

Where is voltage, is current, and is resistance.

We can substitute the given values from the question into the equation and solve:

$\frac{10}{2}=I$

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

Only the change in potential

Only the path traveled

Both the potential and the path traveled

Neither the potential nor the path traveled

Only the magnetic field

Explanation

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.

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?

$\frac{1}{2}C$

$\frac{1}{C}$

$C^{2}$

Explanation

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.

$\frac{1}{2}C$

$\frac{1}{C}$

$C^{2}$

Explanation

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.

If a circuit contains a battery and a $2\Omega$ resistor what is the current of the circuit?

$\frac{1}{5}A$

Explanation

Ohm's law states:

Where is voltage, is current, and is resistance.

We can substitute the given values from the question into the equation and solve:

$\frac{10}{2}=I$

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?

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

$t = v^{2} \sin \frac{\theta }{g}$

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

$t = -v^{2} \sin \frac{\theta }{g}$

$t = v \cos \frac{\theta }{2g}$

Explanation

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: