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givi [52]
4 years ago
9

A space station of diameter 20.0 meters is turning about its axis to simulate gravity at its center rim. How fast must it rotate

to produce an outer rim acceleration of 9.80 m/s^2 ?
Physics
1 answer:
bezimeni [28]4 years ago
4 0

Answer:

9.89 m/s

Explanation:

d = diameter of the space station = 20.0 m

r = radius of the space station

radius of the space station is given as

r = (0.5) d

r = (0.5) (20.0)

r = 10 m

a = acceleration produced at outer rim = 9.80 m/s²

v = speed at which it rotates

acceleration is given as

a = \frac{v^{2}}{r}

9.80 = \frac{v^{2}}{10}

v = 9.89 m/s

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Answer:

1keff=1k1+1k2

see further explanation

Explanation:for clarification

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?

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also for effective force constant

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