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High School Physics : High School Physics

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Example Questions

Example Question #681 : High School Physics

Why is it that photons can move at the speed of light, but other particles cannot?

Possible Answers:

Photons do not have mass

Photons are similar in size to electrons

Photons lack electrical charge

Photons contain more energy than other particles

Correct answer:

Photons do not have mass

Explanation:

Photons can move at the speed of light because they have no mass. Anything with mass can not be accelerated to the speed of light—it takes an infinite amount of energy to do so. We can get very close, but never exactly there.

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Example Question #1 : Electricity And Magnetism

Calculate the magnitude of the electric field at a point that is located $\small 15cm$ directly north of a point charge, $\small Q = -15.0 * 10^{-6}C$ .

$k=9*10^9\frac{m^2N}{C^2}$

Possible Answers:

$-6.0*10^6\frac{N}{C}$

$3.0*10^5\frac{N}{C}$

$-6.0*10^2\frac{N}{C}$

$10.0*10^6\frac{N}{C}$

$-3.0*10^2\frac{N}{C}$

Correct answer:

$-6.0*10^6\frac{N}{C}$

Explanation:

The magnitude of an electric field due to a single point charge is given by the formula:

$E=k\frac{Q}{r^2}$

The variable $\small k$ refers to Coulomb's constant. We are given the charge and the distance between points. To use the equation, first convert the distance to meters.

$15cm*\frac{1m}{100cm}=0.15m$

We can now use the given constant, charge, and distance in the equation to solve for the magnetic field strength.

$E=(9*10^9\frac{m^2N}{C^2})(\frac{-15.0*10^{-6}C}{(0.15)^2})$

$E=(9*19^9\frac{m^2N}{C^2})(-6.67*10^{-4}\frac{C}{m^2})$

$E=-6.0*10^6\frac{N}{C}$

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Example Question #1 : Electric Circuits

What is the voltage of a circuit with 10A of current and $20\Omega$ of resistance?

Possible Answers:

200V

30V

0.5V

10V

Correct answer:

200V

Explanation:

For this problem, use Ohm's law: V=IR . In this equation is the voltage, is the current, and is the resistance.

Plug in the given values and solve for the voltage.

$V=10A* 20\Omega$

V=200V

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Example Question #2 : Electricity And Magnetism

What is the current of a circuit with a voltage of 12V and a total resistance of $5\Omega$ ?

Possible Answers:

4.8A

0.42A

2.4A

60A

Correct answer:

2.4A

Explanation:

For this problem, use Ohm's law: V=IR . In this equation is the voltage, is the current, and is the resistance.

We can re-arrange the equation to solve specifically for .

$\frac{V}{R}=I$

Plug in the given values for voltage and resistance to solve for the current.

$\frac{V}{R}=I$

$\frac{12V}{5\Omega}=I$

2.4A=I

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Example Question #3 : Electric Circuits

What is the resistance in a circuit with a voltage of 50V and a current of ?

Possible Answers:

$42\Omega$

$0.16\Omega$

$58\Omega$

$400\Omega$

$6.25\Omega$

Correct answer:

$6.25\Omega$

Explanation:

For this problem, use Ohm's law: V=IR . In this equation is the voltage, is the current, and is the resistance.

We can re-arrange the equation to solve specifically for .

$\frac{V}{I}=R$

Plug in the given values for voltage and current and solve for resistance.

$\frac{V}{I}=R$

$\frac{50V}{8A}=R$

$6.25\Omega=R$

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Example Question #4 : Electric Circuits

A battery is connected to a circuit. The measured current is . What is the equivalent resistance?

Possible Answers:

$4.5\Omega$

$18\Omega$

$0.22\Omega$

$9\Omega$

$10\Omega$

Correct answer:

$4.5\Omega$

Explanation:

For this problem, use Ohm's law: V=IR .

We are given the voltage and current, allowing us to solve for the resistance.

9V=2A* R

$\frac{9V}{2A}=R$

$4.5\Omega=R$

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Example Question #3 : Electricity And Magnetism

A closed electric circuit is set up so that there is a current of 9.9A and a voltage of 30.22V . What is the resistance in the circuit?

Possible Answers:

$299.18\Omega$

$20.32\Omega$

$6.1\Omega$

$3.05\Omega$

$0.33\Omega$

Correct answer:

$3.05\Omega$

Explanation:

For this problem, use Ohm's law: V=IR .

We are given the current and the voltage. Using these terms, we can solve for the resistance.

V=IR

30.22V=(9.9A) R

$\frac{30.22V}{9.9A}=R$

$3.05\Omega=R$

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Example Question #4 : Electricity And Magnetism

The current in a circuit is 20A . If the voltage is 35V , what is the total resistance?

Possible Answers:

$3.5\Omega$

$0.57\Omega$

$1.75\Omega$

$70\Omega$

$700\Omega$

Correct answer:

$1.75\Omega$

Explanation:

For this problem use Ohm's law:

V=IR

We are given the current and the voltage, allowing us to solve for the resistance.

35V=(20A)*R

$R=\frac{35V}{20A}$

$R=1.75\Omega$

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Example Question #5 : Electric Circuits

An electrical circuit has a total of $3.22\Omega$ resistance and . What is the current?

Possible Answers:

25.76A

0.403A

11.22A

2.48A

0.089A

Correct answer:

2.48A

Explanation:

For this problem use Ohm's law:

V=IR

We are given the resistance and the voltage, allowing us to solve for the current.

$8V=I*(3.22\Omega)$

$I=\frac{8V}{3.22\Omega}$

I=2.48A

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Example Question #6 : Electric Circuits

An electrical circuit has a current of and $32\Omega$ of resistance. What is the voltage?

Possible Answers:

0.25V

128V

256V

40V

Correct answer:

256V

Explanation:

For this problem use Ohm's law:

V=IR

We are given the resistance and the current, allowing us to solve for the voltage.

$V=(8A)(32\Omega)$

V=256V

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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #681 : High School Physics

Example Question #1 : Electricity And Magnetism

Example Question #1 : Electric Circuits

Example Question #2 : Electricity And Magnetism

Example Question #3 : Electric Circuits

Example Question #4 : Electric Circuits

Example Question #3 : Electricity And Magnetism

Example Question #4 : Electricity And Magnetism

Example Question #5 : Electric Circuits

Example Question #6 : Electric Circuits

All High School Physics Resources

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