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elena-14-01-66 [18.8K]

3 years ago

12

What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?

1 answer:

prohojiy [21]3 years ago

7 0

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about

(1 / √0.66) = 1.23 times

the Earth's radius, so the height is about 910 miles above the surface.

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3 years ago

Last week we investigated the black hole at the center of our galaxy, with massMX= 8.5×1036kg. An object whose size is on the or

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Answer:

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erastova [34]

Answer:

(a). The ball's centripetal acceleration is $16.17\times10^{2}\ m/s^2$

(b). The magnitude of the net force is 232.9 N.

Explanation:

Given that,

Mass of baseball = 144 g

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Using formula of centripetal acceleration

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Put the value into the formula

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(b). We need to calculate the magnitude of the net force that is acting on the ball just before it is released

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$F=\dfrac{144\times10^{-3}\times(36.2)^2}{81\times10^{-2}}$

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(b). The magnitude of the net force is 232.9 N.

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