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Maurinko [17]
3 years ago
10

g A ping pong ball (thin shell sphere) rolls down an incline at 30° from rest. What is its acceleration

Physics
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

Explanation:

According to newtons second law of motion;

F = ma .... 1

Also the force acting aong the inclines is expressed as;

F = mgsintheta

m is the mass of the object

a is the acceleration

theta is angle of inclination

Equate 1 and 2

ma = mg sin theta

a = gsin(theta)

a = 9.8sin30

a = 9.8(0.5)

a = 4.9m/s²

Hence the acceleration of the ping pong is 4.9m/s²

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Answer:

The ball has an acceleration of -380 m/s², this means the ball slows down

An acceleration of -380 m/s² is the equivalent of 38.736 g's

Explanation:

Step 1: Data given

Velocity of the baseball at time t=0 = 38 m/s

At time t, the ball stops. This means v = 0

time before stops = 0.1s

Step 2: Calculate the acceleration

v= v0+at

with v= the velocity of the ball at time t = 0. v= 0

with v0 = the velocity of the ball at time t=0. v0 = 38 m/s

with a= the acceleration in m/s²

with t = time in seconds

0 = 38 + a*0.1

a = -380 m/s²

The ball has an acceleration of -380 m/s², this means the ball slows down

An acceleration of -380 m/s² is the equivalent of 38.736 g's

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3 years ago
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Answer:

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