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lora16 [44]
3 years ago
14

The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it

s north-south spin axis is 7.292(10-5) rad/s.Use the value for the radius of the earth as 6371 km and determine the velocities of points A, B, C, and D, all of which are on the equator. The inclination of the axis of the earth is neglected.
Physics
1 answer:
Sindrei [870]3 years ago
3 0

Answer with Explanation:

We are given that

Speed ,v_0=107257 kmph

Angular velocity,\omega=7.292\times 10^{-5} rad/s

Radius of earth,r=6371 km=6371000 m

1 km=1000 m

Linear velocity,v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s

Linear velocity,v=464.57\times \frac{18}{5}=1672.46 km/h

Velocity at point A,v_A=-vi+v_0 j=-1672.46 i+107257j kmph

Velocity at point B,v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph

Velocity at point C,v_C=v_0j+vi=1672.46 i+107257j kmph

Velocity at point  D,v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph

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Both the astronauts and photographer have the same displacement

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Answer:

Part a)

h' = \frac{10}{14} h

Part b)

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Part c)

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Explanation:

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now on the right side of the bowl there is no friction

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so we have

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh

\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh

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h' = \frac{10}{14} h

Part b)

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Part c)

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so on smooth it will go to lower height

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