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High School Physics : Motion and Mechanics

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Example Questions

Example Question #23 : Circular Motion

The torque applied to a wrench is 16N*m . If the force applied to the wrench is 40N , how long is the wrench?

Possible Answers:

56m

24m

2.5m

0.4m

Correct answer:

0.4m

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the total torque and the force applied. Using these values, we can solve for the length of the wrench.

16N*m=(40N)r

$\frac{16N*m}{40N}=r$

0.4m=r

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Example Question #171 : Motion And Mechanics

If the force applied to a 0.21m wrench is 50N , how much torque is generated?

Possible Answers:

21N*m

238.1N*m

0.0042N*m

10.5N*m

50.21N*m

Correct answer:

10.5N*m

Explanation:

The formula for torque is:

$\tau=F*r$

We are given the values for the force and the length. Using these values, we can multiply to find the torque.

$\tau=50N*0.21m$

$\tau=10.5N*m$

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Example Question #172 : Motion And Mechanics

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:

28N

32N

14.5N

13.5N

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 #173 : Motion And Mechanics

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

Possible Answers:

3.9N

1.26N

14N

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 #174 : Motion And Mechanics

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:

$31\frac{m}{s}$

$7.44\frac{m}{s}$

$129.17\frac{m}{s}$

$1.79\frac{m}{s}$

$29.76\frac{m}{s}$

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 #175 : Motion And Mechanics

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:

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

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

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

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

There is insufficient information to solve

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 #176 : Motion And Mechanics

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 #22 : Using Circular Motion Equations

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:

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

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

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

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

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

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 #178 : Motion And Mechanics

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

Possible Answers:

13.2N*m

12N*m

12.1N*m

14.52N*m

15.97N*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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High School Physics : Motion and Mechanics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #23 : Circular Motion

Example Question #171 : Motion And Mechanics

Example Question #172 : Motion And Mechanics

Example Question #173 : Motion And Mechanics

Example Question #174 : Motion And Mechanics

Example Question #175 : Motion And Mechanics

Example Question #176 : Motion And Mechanics

Example Question #22 : Using Circular Motion Equations

Example Question #177 : Motion And Mechanics

Example Question #178 : Motion And Mechanics

All High School Physics Resources

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