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icang [17]
3 years ago
9

An object is 45 m above the ground when it is dropped. How fast is the object going just before it hits the ground?

Physics
1 answer:
leonid [27]3 years ago
3 0
This is a kinematics question.
v^{2} = v_{0} ^{2} + 2g(y - y_{0}) \\ 
v^{2} = 0 + 2(-9.8)(0 - 45) \\ 
v^{2} = 882 \\ v =  \sqrt{882}  = 29.7 m/s

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Which theory suggests that the moon and the earth formed at the same time from dust and the solar nebula that formed the sun?
ad-work [718]

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6 0
3 years ago
A 70- kg bicycle rides his 9.8- kg bicycle with a speed of 16 m/ s. What is the magnitude of the braking force of the bicycle co
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Explanation:

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3 0
3 years ago
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Ilia_Sergeevich [38]

Answer:

A: 1.962

B: 3.924

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g = G *M /R^2

g = 9.807*M/R^2 the gravitational constant of ground level on earth is about 9.807

g = 9.807*5lbs/R^2 the average brick is about 5 pounds.

g = 9.807*5*10^2.   I'm assuming the height is around ten feet to help you out.

with these numbers plugged in you get an acceleration of 0.4905 a final velocity after 4 seconds 1.962. It's height fallen after 4 seconds is 3.924.

( M = whatever the brick weighs it's not specified in the question)

(R = the distance from the ground or how high the scaffold is)

(hopefully you can just plug your numbers in there hope this helps)

6 0
3 years ago
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