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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

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6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Using Circular Motion Equations

The torque applied to a wrench is 14N*m . If the length of the wrench is 0.5m , how much force is applied to it?

Possible Answers:

13.5N

32N

28N

14.5N

Correct answer:

28N

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the total torque and the length of the wrench. Given that the pivot point will be at one end of the wrench, and the force will be applied to the other end, this length can be used as our radius.

14N*m=F*0.5m

$\frac{14N*m}{0.5m}=F$

28N=F

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Example Question #174 : Motion And Mechanics

How much force is required for a 0.3m hammer to produce 4.2N*m of torque?

Possible Answers:

14N

3.9N

1.26N

45N

Correct answer:

14N

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the length of the hammer (radius of the swing) and the torque produced. Using these values, we can solve for the force required.

4.2N*m=F(0.3m)

$\frac{4.2N*m}{0.3m}=F$

14N=F

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Example Question #31 : Circular Motion

A top spins so that its angular velocity is $\small 31\frac{rad}{s}$ . If the top has a radius of 0.24m , what is the tangential velocity of a point on the edge of the top?

Possible Answers:

$129.17\frac{m}{s}$

$7.44\frac{m}{s}$

$29.76\frac{m}{s}$

$1.79\frac{m}{s}$

$31\frac{m}{s}$

Correct answer:

$7.44\frac{m}{s}$

Explanation:

The relationship between tangential velocity and angular velocity is:

$\omega=\frac{v}{r}$

We are given the angular velocity and the radius, allowing us to solve for the linear (or tangential) velocity.

$\omega=\frac{v}{r}$

$31\frac{rad}{s}=\frac{v}{0.24\frac{m}{rad}}$

$0.24\frac{m}{rad}*31\frac{rad}{s}=v$

$7.44\frac{m}{s}=v$

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Example Question #35 : Circular Motion

A top spins so that the angular velocity is $\small 31\frac{rad}{s}$ . If the top has a radius of 0.24m , what is the centripetal acceleration acting on a point on the edge of the top?

Possible Answers:

$4004.17\frac{m}{s^2}$

$3.19\frac{m}{s^2}$

$230.64\frac{m}{s^2}$

There is insufficient information to solve

$13.28\frac{m}{s^2}$

Correct answer:

$230.64\frac{m}{s^2}$

Explanation:

The formula for centripetal acceleration is:

$a_c=\frac{v^2}{r}$

We know the radius, but we need to find the tangential velocity. The relationship between tangential velocity and angular velocity is:

$\omega=\frac{v}{r}$

We are given the angular velocity and the radius, allowing us to solve for the linear (or tangential) velocity.

$31\frac{rad}{s}=\frac{v}{0.24\frac{m}{rad}}$

$0.24\frac{m}{rad}*31\frac{rad}{s}=v$

$7.44\frac{m}{s}=v$

Now we have the linear velocity and the radius, allowing us to use the first equation to find the centripetal acceleration.

$a_c=\frac{(7.44\frac{m}{s})^2}{0.24m}$

$a_c=\frac{55.35\frac{m^2}{s^2}}{0.24m}$

$a_c=230.64\frac{m}{s^2}$

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Example Question #241 : High School Physics

A top spins so that the angular velocity is $\small 31\frac{rad}{s}$ . If the top has a radius of 0.24m , what is the centripetal force acting on a point on the edge of the top?

Possible Answers:

3.19N

55.35N

230.64N

961N

There is insufficient information to solve

Correct answer:

There is insufficient information to solve

Explanation:

The formula for centripetal force is:

$F_c=ma_c=\frac{mv^2}{r}$

While we can find what the acceleration will be, there is no way for us to find the mass. The force cannot be calculated.

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Example Question #176 : Motion And Mechanics

A girl sits on the edge of a merry-go-round. If she travels at $0.02\frac{m}{s}$ and the radius of her motion is 2.1m , what is the centripetal acceleration acting upon her?

Possible Answers:

$0.0004\frac{m}{s^2}$

$0.00019\frac{m}{s^2}$

$0.019\frac{m}{s^2}$

$0.00036\frac{m}{s^2}$

$5.26\frac{m}{s^2}$

Correct answer:

$0.00019\frac{m}{s^2}$

Explanation:

The formula for centripetal acceleration is:

$a_c=\frac{v^2}{r}$

We are given values for the linear velocity and radius, allowing us to solve for the centripetal acceleration.

$a_c=\frac{(0.02\frac{m}{s})^2}{2.1m}$

$a_c=\frac{0.0004\frac{m^2}{s^2}}{2.1m}$

$a_c=0.00019\frac{m}{s^2}$

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Example Question #177 : Motion And Mechanics

A 41kg girl sits on the edge of a merry-go-round. If she travels at $0.02\frac{m}{s}$ and the radius of her motion is 2.1m , what is the centripetal force acting upon her?

Possible Answers:

0.016N

210.5N

0.0078N

0.661N

1.244N

Correct answer:

0.0078N

Explanation:

The formula for centripetal force is:

$F_c=\frac{mv^2}{r}$

We are given the girl's mass, her linear velocity, and the radius of her motion. Using these values, we can calculate the centripetal force.

$F_c=\frac{(41kg)(0.02\frac{m}{s})^2}{2.1m}$

$F_c=\frac{(41kg)(0.0004\frac{m^2}{s^2})}{2.1m}$

F_c=0.0078N

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Example Question #177 : Motion And Mechanics

How much torque is generated by applying 13.2N of force to the end of a 1.1m lever?

Possible Answers:

12N*m

15.97N*m

12.1N*m

14.52N*m

13.2N*m

Correct answer:

14.52N*m

Explanation:

The equation for torque is:

$\tau=F*r$

We are given the force and distance (radius), allowing us to solve for the torque by multiplication.

$\tau=13.2N*1.1m$

$\tau=14.52N*m$

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Example Question #181 : Motion And Mechanics

If 16N*m of torque is generated by applying 18N of force, how long is the lever arm?

Possible Answers:

0.89m

0.5m

1.13m

There is insufficient information to solve

Correct answer:

0.89m

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the force applied and the torque generated, allowing us to solve for the length of the lever arm.

16N*m=(18N)r

$\frac{16N*m}{18N}=r$

0.89m=r

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Example Question #41 : Circular Motion

If 11N*m of torque is generated using a 0.43m wrench, how much force is required?

Possible Answers:

0.04N

17.74N

25.58N

11.43N

12.79N

Correct answer:

25.58N

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the length of the lever arm and the torque generated, allowing us to solve for the force applied.

11N*m=F(0.43m)

$\frac{11N*m}{0.43m}=F$

25.58N

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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 #11 : Using Circular Motion Equations

Example Question #174 : Motion And Mechanics

Example Question #31 : Circular Motion

Example Question #35 : Circular Motion

Example Question #241 : High School Physics

Example Question #176 : Motion And Mechanics

Example Question #177 : Motion And Mechanics

Example Question #177 : Motion And Mechanics

Example Question #181 : Motion And Mechanics

Example Question #41 : Circular Motion

All High School Physics Resources

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