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High School Physics : 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 #162 : Motion And Mechanics

A 2kg ball rolls around the edge of a circle with a radius of . If there is a 12N centripital force acting upon it, what is its speed?

Possible Answers:

$6\frac{m}{s}$

$2.45\frac{m}{s}$

$3\frac{m}{s}$

$3.99\frac{m}{s}$

$1.82\frac{m}{s}$

Correct answer:

$2.45\frac{m}{s}$

Explanation:

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is F_c=ma_c .

Since we know the mass and the force, we can find the accleration.

F_c=ma_c

12N=(2kg)a_c

$\frac{12N}{2kg}=a_c$

$6\frac{m}{s^2}=a_c$

Centripetal acceleration is given by the formula $a_c=\frac{v^2}{r}$ , where is the perceived tangential velocity and is the radius of the circle.

Plug in the given values and solve for the velocity.

$6\frac{m}{s^2}=\frac{v^2}{(1m)}$

$1m* 6\frac{m}{s^2}=v^2$

$6\frac{m^2}{s^2}=v^2$

$\sqrt{6\frac{m^2}{s^2}}=\sqrt{v^2}$

$2.45\frac{m}{s}=v$

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Example Question #163 : Motion And Mechanics

The radius of the sun is 696,000km. If its period is 587.28 hours, what is its tangential velocity at the equator?

Possible Answers:

$700000\pi\frac{km}{hr}$

$2370.25\pi\frac{km}{hr}$

$14446.36\pi\frac{km}{hr}$

$98101.23\frac{km}{hr}$

$589.22\frac{km}{hr}$

Correct answer:

$2370.25\pi\frac{km}{hr}$

Explanation:

The equation for velocity is $v=\frac{\Delta x}{\Delta t}$ . Using the circumference of the circle as the distance and the time as the period, we can rewrite the equation for velocity: $v=\frac{2\pi r}{T}$ .

Plug in the given values and solve for the velocity.

$v=\frac{2\pi (696,000km)}{587.28hr}$

$v=\frac{\pi*1392000km}{587.28hr}$

$v=2370.25\pi\frac{km}{hr}$

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Example Question #164 : Motion And Mechanics

A large burst of wind exerts 750N of force on a tree 10m tall. How much torque is exerted on the tree? (Assume that the entire force is pointed at the top of the tree, the pivot point is at the ground, and ignore center of mass fluctuations)

Possible Answers:

7.5N*m

75000N*m

7500N*m

75N*m

740N*m

Correct answer:

7500N*m

Explanation:

The formula for torque is:

$\tau=F*r$

Plug in our given values:

$\tau=750N*10m$

$\tau=7500N*m$

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Example Question #4 : Using Circular Motion Equations

of force is exerted on the end of a 0.23m wrench. How much torque is generated? (Ignore any center of mass fluctuations.)

Possible Answers:

34.78N*m

1.84N*m

0.92N*m

8.23N*m

7.77N*m

Correct answer:

1.84N*m

Explanation:

The formula for torque is:

$\tau=F*r$

Plug in our given values:

$\tau=8N*0.23m$

$\tau=1.84N*m$

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Example Question #166 : Motion And Mechanics

The velocity of a certain ferris wheel is $6\frac{m}{s}$ . If the wheel has a diameter of 12m , what is the centripetal acceleration generated by the wheel?

Possible Answers:

$1\frac{m}{s^2}$

$36\frac{m}{s^2}$

$30\frac{m}{s^2}$

$3\frac{m}{s^2}$

$6\frac{m}{s^2}$

Correct answer:

$6\frac{m}{s^2}$

Explanation:

The formula for centripetal acceleration is $a_c=\frac{v^2}{r}$ .

The problem gives us a diamater. Since radius is defined as half of an object's diameter, the radius of the ferris wheel is .

Now we can plug in and solve:

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

$a_c=\frac{(6\frac{m}{s})^2}{6m}$

$a_c=\frac{36\frac{m^2}{s^2}}{6m}$

$a_c=6\frac{m}{s^2}$

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Example Question #161 : Motion And Mechanics

An object moves in a circle with a constant velocity of $3\frac{m}{s}$ . If the radius of the circle is , what is the centripetal acceleration on the object?

Possible Answers:

$9\frac{m}{s^2}$

$4.5\frac{m}{s^2}$

$6\frac{m}{s^2}$

$18\frac{m}{s^2}$

$1.5\frac{m}{s^2}$

Correct answer:

$4.5\frac{m}{s^2}$

Explanation:

The formula for centripetal acceleration is:

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

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

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

$a_c=\frac{(3\frac{m}{s})^2}{2m}$

$a_c=\frac{9\frac{m^2}{s^2}}{2m}$

$a_c=4.5\frac{m}{s^2}$

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Example Question #168 : Motion And Mechanics

If the centripetal force on a 3.3kg object is 14.5N , and the object is 1.9m from the center of the circle, what is its centripetal acceleration?

Possible Answers:

$3.3\frac{m}{s^2}$

$5.6\frac{m}{s^2}$

$11.2\frac{m}{s^2}$

$0.443\frac{m}{s^2}$

$4.43\frac{m}{s^2}$

Correct answer:

$4.43\frac{m}{s^2}$

Explanation:

The formula for centripetal force is F_c=ma_c .

We know the value of the force, as well as the mass of the object. Using these values, we can solve for the acceleration; the radius is extraneous information.

F_c=ma_c

14.5N=3.3kg*a_c

$\frac{14.5N}{3.3kg}=a_c$

$4.34\frac{m}{s^2}=a_c$

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Example Question #162 : Motion And Mechanics

If the centripetal force on a 3.3kg object is 14.5N , and the object is 1.9m from the center of the circle, what is its tangential velocity?

Possible Answers:

$8.25\frac{m}{s}$

$1.51\frac{m}{s}$

$2.083\frac{m}{s}$

$4.34\frac{m}{s}$

$2.87\frac{m}{s}$

Correct answer:

$2.87\frac{m}{s}$

Explanation:

The relationship between centripetal acceleration and tangential velocity is:

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

We can find the centripetal acceleration using the centripetal force and the mass.

The formula for centripetal force is F_c=ma_c .

Using the force and mass from the question, we can solve for the acceleration.

F_c=ma_c

14.5N=3.3kg*a_c

$\frac{14.5N}{3.3kg}=a_c$

$4.34\frac{m}{s^2}=a_c$

Now that we know the acceleration, we can return to the first equation. Using the acceleration and the radius from the question, we can solve for the velocity.

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

$4.34\frac{m}{s^2}=\frac{v^2}{1.9m}$

$4.34\frac{m}{s^2}*1.9m=v^2$

$8.246\frac{m^2}{s^2}=v^2$

$\sqrt{8.246\frac{m^2}{s^2}}=\sqrt{v^2}$

$2.87\frac{m}{s}=v$

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Example Question #171 : Motion And Mechanics

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:

2.5m

24m

0.4m

56m

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 #172 : Motion And Mechanics

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

Possible Answers:

10.5N*m

238.1N*m

50.21N*m

21N*m

0.0042N*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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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #162 : Motion And Mechanics

Example Question #163 : Motion And Mechanics

Example Question #164 : Motion And Mechanics

Example Question #4 : Using Circular Motion Equations

Example Question #166 : Motion And Mechanics

Example Question #161 : Motion And Mechanics

Example Question #168 : Motion And Mechanics

Example Question #162 : Motion And Mechanics

Example Question #171 : Motion And Mechanics

Example Question #172 : Motion And Mechanics

All High School Physics Resources

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