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Alekssandra [29.7K]
3 years ago
7

How far will a freely falling object fall from rest in six seconds?

Physics
2 answers:
bija089 [108]3 years ago
5 0
Tabulate, then find the proper equation, then solve.

vi=0
vf=?
t=6s
a=g=9.8m/s²

vf=vi+at
vf=gt
vf= (find this yourself)
navik [9.2K]3 years ago
4 0
Here are two ways to do it:

<u>Way #1:</u>
-- You know that gravity accelerates things that are falling freely, adding 9.8 m/s to their speed every second.

-- After 6 seconds, an object that fell from rest winds up falling at (9.8 x 6) = 58.8m/s.

-- During that time, its average speed was 1/2(0 + 58.8) = 29.4 m/s .

-- In 6 seconds, at an average speed of 29.4 m/s, it covers (29.4 x 6) = <em>176.4meters</em>.

<u>Way #2:</u>
This way only works if you remember the formula for the distance covered during uniform acceleration.

             D  =  1/2 (acceleration) (time)²

Acceleration = gravity = 9.8 m/s²
Time  =  6 seconds

           D  =  1/2 (9.8) (6)²  =  <em>176.4 meters</em>.
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A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s
sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

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distance=0.124 m

Explanation:

mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\   v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\   L=0.124m

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