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Rashid [163]
3 years ago
6

A playground merry-go-round with a radius of 2.0 m and a rotational inertia of 100 kg m2 is rotating at 3.0 rad/s. A child with

a mass of 22 kg jumps onto the edge of the merry-go-round, traveling radially inward. What is the new angular speed of the merry-go-round
Physics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

Explanation:

Conservation of angular momentum

The MGR originally has momentum

L = 100(3.0) = 300 kg•m²/s²

The child can be thought of as a point mass with I = mr²

When she jumps onto the rim of the MGR

300 = (100 + 22(2.0²)ω

ω = 300 / 188 = 1.5957... 1.6 rad/s

As she moves toward the center of the MGR, her moment of inertia goes to zero as her radius goes to zero.

The angular velocity when she reaches the center will again be 3.0 rad/s

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A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction
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consider the motion in x-direction

v_{ox} = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

a_{x} = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = v_{ox} t + (0.5) a_{x} t²

100 =  v_{ox} (4.60)

v_{ox} = 21.7 m/s


consider the motion along y-direction

v_{oy} = initial velocity in y-direction = ?

Y = vertical displacement  = 0 m

a_{y} = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = v_{oy} t + (0.5) a_{y} t²

0 = v_{oy} (4.60) + (0.5) (- 9.8) (4.60)²

v_{oy} = 22.54 m/s

initial velocity is given as

v_{o} = sqrt((v_{ox})² + (v_{oy})²)

v_{o} = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

6 0
3 years ago
if we ignore air resistance the mass of an object does not affect the rate at which it accelerate why?​
quester [9]

Answer:

See explanation

Explanation:

The acceleration due to gravity on an object is independent of the mass of the object. This is so because, the acceleration due to gravity depends only on the radius of the earth and the mass of the earth.

As a result of this, all objects are accelerated to the same extent and should reach the ground at the same time when released from a height as long as other forces other than gravity are not at work.

5 0
3 years ago
When an object is moving away from you while making a sound, the pitch of the sound is perceived to be ___________. lower unchan
Len [333]

the answer is higher

8 0
3 years ago
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A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with
beks73 [17]

Answer:

4.6 kHz

Explanation:

The formula for the Doppler effect allows us to find the frequency of the reflected wave:

f'=(\frac{v}{v-v_s})f

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v is the speed of sound

vs is the speed of the wave source

In this problem, we have

f = 41.2 kHz

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vs = 33.0 m/s

Therefore, if we substitute in the equation we find the frequency of the reflected wave:

f'=(\frac{330 m/s}{330 m/s-33.0 m/s})(41.2 kHz)=45.8 kHz

And the frequency of the beats is equal to the difference between the frequency of the reflected wave and the original frequency:

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