PHYSICS COLLEGE

A 24.1 N solid sphere with a radius of 0.151 m is released from rest and rolls, without slipping, 1.7 m down a ramp that is inclined at 34o above the horizon. What is the total kinetic energy of the sphere at the bottom of the ramp?What is the angular speed of the sphere at the bottom of the ramp? How many radians did the sphere rotate through as it rolled down the ramp What was the angular acceleration of the sphere as it rolled down the ramp

Answers

Answer 1

Answer:

Answer

given,

weight of solid sphere = 24.1 N

m = 24.1/g = 24.1/10 = 2.41 Kg

radius = R = 0.151 m

height of the ramp = 1.7 m

angle with horizontal = 34°

acceleration due to gravity = 10 m/s²

using energy conservation

$(1)/(2)I\omega^2 + (1)/(2)mv^2 = mgh$

I for sphere

$I = (2)/(5)mr^2$ v = r ω

$(1)/(2)\ (2)/(5)mr^2* (v^2)/(r^2) + (1)/(2)mv^2 = mgh$

$(7)/(10)mv^2 = mgh$

$h = (0.7 v^2)/(g)$

$v = \sqrt{(h * g)/(0.7)}$

$v = \sqrt{(1.7 * 10)/(0.7)}$

v = 4.93 m/s

b) rotational kinetic energy

$KE=(1)/(2)I\omega^2$

$KE=(1)/(2)\ (2)/(5)mr^2* (v^2)/(r^2)$

$KE=(1)/(5)mv^2$

$KE=(1)/(5)* 2.41 * 4.93^2$

KE = 11.71 J

c) Translation kinetic energy

$KE=(1)/(2)mv^2$

$KE=(1)/(2)* 2.41 \time 4.93^2$

$KE=29.28\ J$

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A string is attached to the rear-view mirror of a car. A ball is hanging at the other end of the string. The car is driving around in a circle, at a constant speed. Which of the following lists gives all of the forces directly acting on the ball?a. tension
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d. tension, gravity, the centripetal force, and friction

Answers

Final answer:

The forces directly acting on the ball hanging from a rear-view mirror while a car drives in a circle are tension, gravity, and the centripetal force.

Explanation:

The correct answer is c. tension, gravity, and the centripetal force.

When the car is driving in a circle, the ball experiences both tension and gravity. The tension in the string is what keeps the ball from falling, while gravity pulls the ball downward.

In addition to tension and gravity, the ball also experiences the centripetal force. This force is directed towards the center of the circular motion and keeps the ball moving in a circular path.

Learn more about Forces acting on a ball in circular motion here:

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Consider the following statements. A. Heat flows from an object at higher temperature to an object at lower temperature; B. Heat flows from an object in liquid state to an object in solid state; C. Heat flows from an object with higher thermal energy to one with lower thermal energy. Which statements are true? 1. A only 2. A and C only 3. B and C only 4. None is true.

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Only ' A ' is always true. (choice-1)

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What is the correct formula for barium nitride? A. Ba3N2,
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Answers

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Part 1) A cop car traveling at 25 m/s has a siren producing a frequency of 700 Hz. A felon jumps on his motorcycle and speed off in the opposite direction of 15 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?Part 2) The cop does a U-turn and speeds towards the felon at 30 m/s, while the felon speeds up to 20 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?
Part 3) What if the felon then sped up to 30 m/s and all other conditions remained the same?

Answers

1) 621.8 Hz

2) 719.3 Hz

3) 700 Hz

Explanation:

The Doppler effect occurs when there is a source of a wave in relative motion with respect to an observer.

When this happens, the frequency of the wave appears shifted to the observer, according to the equation:

$f'=(v\pm v_o)/(v \pm v_s)f$

where

f is the real frequency of the sound

f' is the apparent frequency of the sound

v is the speed of the sound wave

$v_o$ is the velocity of the observer, which is negative if the observer is moving away from the source, positive if the observer is moving towards the source

$v_s$ is the velocity of the source, which is negative if the source is moving towards the observer, positive if the source is moving away

In this problem we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=-25 m/s$ is the velocity of the car with the siren

$v_o = +15 m/s$ is the velocity of the felon (he's moving away from the siren)

So, the frequency heard by the felon is

$f=(343-25)/(343+15)(700)=621.8 Hz$

In this case, the cop does a U-turn and speeds towards the felon at 30 m/s.

This means that now the siren is moving towards the observer (so, $v_s$ becomes positive), while the sign of $v_o$ still remains positive.

So we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=+30 m/s$ is the velocity of the car with the siren

$v_o = +20 m/s$ is the velocity of the felon

So, the frequency heard by the felon is

$f=(343+30)/(343+20)(700)=719.3 Hz$

In this case, the felon speeds up to 30 m/s.

This means that now the felon and the siren are moving with the same relative velocity: so, it's like they are not moving relative to each other, so the frequency will not change.

In fact we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=+30 m/s$ is the velocity of the car with the siren

$v_o = +30 m/s$ is the velocity of the felon

So, the frequency heard by the felon is

$f=(343+30)/(343+30)(700)=700 Hz$

So, the frequency will not change.

A heavy object and a light object are dropped from the same height. If we neglect air resistance, which will hit the ground first?

Answers

Answer:

None, both objects will hit ground at the same time.

Explanation:

Assuming no air resistance present, and that both objects start from rest, we can apply the following kinematic equation for the vertical displacement:

$\Delta h = (1)/(2)*g*t^(2) (1)$

As the left side in (1) is the same for both objects, the right side will be the same also.
Since g is constant close to the surface of the Earth, it's also the same for both objects.
So, the time t must be the same for both objects also.

A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train

Answers

Answer:

Apparent frequency of the bell to the observer is 546.12 Hz

Explanation:

The frequency of train bell (frequency of source) = 505 Hz

The speed of train (observer) = 27.6 m/s

The speed of sound in the air is (velocity of sound) = 339 m/s

The apparent frequency of the bell to the observer is calculated as follows:

Apparent frequency of bell to the observer.

$= \text{frequency of source} * (Observer + velocity \ of \ sound )/( velocity \ of \ sound ) \n= 505 * (27.6 + 339)/(339) \n= 546.12 Hz$

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