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3 years ago

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Please help!! An astronaut drops a rock on the surface of an asteroid. The rock is released from rest at a height of 0.79 m abov

e the ground, and hits the ground 1.35 s later.
What is the acceleration due to gravity on this asteroid?

1 answer:

enot [183]3 years ago

3 0

Answer:

0.87ms‐²

Explanation:

Explanation is on photo above, it is easier to show steps than explain using words

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The purpose of the defense is to _____.

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The purpose of the defense is to <span>prevent the opposing offense from advancing the ball.

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a runner runs 2.88 m/s north. she accelerates .350 m/s ^2 at a -52.0 degree angle at the point in the motion where she is runnin

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3 years ago

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti

34kurt

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r = 0.89 x 0. 5 = 0.445 m

Angular velocity

$\omega =\frac{72}{0.445}=161.8rad/s$

Frequency

$f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min$

Revolutions per minute = 2.33

b) Centripetal acceleration

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

Here linear velocity = 72 m/s

Radius, r = 0.445 m

Substituting

$a=\frac{72^2}{0.445}=11649.44m/s^2$

Centripetal acceleration = 11649.44m/s²

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