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NARA [144]
3 years ago
5

A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the m

agnitude of the average acceleration of the ball during the time it is in contact with the ground?
Physics
1 answer:
rosijanka [135]3 years ago
3 0

Answer:

a= -0.86 m/s²

The negative sign shows that ball down the ground or moving down

Explanation:

Vf² - Vo² = 2gS

where

Vf = velocity of clay as it hits the ground

Vo = initial velocity of clay = 0

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

S = distance travelled by clay = 15 m

Substituting appropriate values,

Vf² - 0 = 2(9.8)(15)  

Vf = 17.15 m/sec.

Formula to use is,  

V - Vf = aT

where

V = velocity of clay when it stops = 0

Vf = 17.15 m/sec (as determined above)

a = acceleration

T = 20 ms  

Put the values to find acceleration

a=(V-Vf)/T

a=(0-17.15)/20

a= -0.86 m/s²

The negative sign shows that ball down the ground

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<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

<h3><u>Explanation;</u></h3>

The mass of proton m=1.6748 × 10^-27 kg;  

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                          = 2.7 × 10^6 × 1.602 × 10^-19 J;

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