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Serjik [45]
3 years ago
9

Two children are spinning on a miracle round. What is 50 cm from the center and the other is 100 cm from the center. Which child

has the larger linear speed?
Physics
1 answer:
GarryVolchara [31]3 years ago
6 0

Answer:

The one that is 100 cm from the center.

Explanation:

Given that merry-go-round means moving at uniform circular motion or constant angular speed. Hence the person who is closer to the center is moving at a slower speed, while the one far away from the center has a larger linear speed because he has moved a long distance in the same amount of time, despite being on these objects.

To illustrate

A with the 50 cm

B with the 100 cm

Hence, to get speed, we have

A = 2 * pi * 50 / T

B = 2 * pi * 100 / T

And by the explanation of angular speed

We have w = 2*pi/T where w is the angular speed in radians/s

Hence we have

A = 50w

B = 100w

Therefore, the one that is 100 cm from the center has a larger linear speed.

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The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

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v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

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iragen [17]

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Explanation:

First, we calculate the angular speed of the satellite:

\omega = \frac{Arc Length}{Time}\\\\For\ One\ Complete\ Revolution\ around\ Cruton:\\\\\omega = \frac{2\pi\ rad}{5.82\ h}\frac{1\ h}{3600\ s}\\\omega =  3\ x\ 10^{-4} rad/s

Now, we use the formula for the gravitational force. Since gravitational force will be acting as the centrifugal force due to circular motion. Therefore,

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where,

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r = radius of orbit = ?

Therefore,

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