No one answeres physic questions, iveneeded help all morning
Answer:
<h2>50m/s^2</h2>
Explanation:
Step one:
given data
initial velocity u= 0m/s since the ball is at rest
time of contact t= 0.3s
final velocity v=15m/s
Required
acceleration a
from the first law of motion
v=u+at
substitute our given data
15=0+a*0.3
15=0.3a
divide both sides by 0.3
a=15/0.3
a=50m/s
<u>The average acceleration is 50m/s^2</u>
Answer:
The answer is option E
Explanation:
The particles vibrate in a perpendicular direction with respect to the wave motion
Let the mass of 2500 kg car be
and it's velocity be
and the mass of 1500 kg car be
and it's velocity be
.
After the bumping the mass be M and it's velocity be V.
By law of conservation of momentum we have

2500 * 5 + 1500 * 1=4000 * V
V = 14000/4000 = 7/2 = 3.5 m/s
So the velocity of the two-car train = 3.5 m/s