Using the conservation of momentum,
ma*va1 + mb*vb1 = ma*va2 + mb*vb2
Let:
ma = mass of the ball
va = velocity of the ball
mb = mass of the man
vb = velocity of the man
The subscript 1 is known as initials while 2 is for finals.
Before the man throws the ball, he starts at rest, meaning the initial velocity of the ball and the initial velocity of the man are zero. So
0 = ma*va2 + mb*vb2
Given ma = 10 kg; va = 20 m/s; mb = 90 kg; vb is unknown, therefore
-(mb*vb2) = ma*va2
vb2 = -(ma*va2)/mb2 = -(10*20)/90 = -2.22 m/s
Notice that his velocity is negative because when he finally throws the ball (say to the right), he moves at the opposite direction (that is to the left) on which he stands on the frictionless surface.
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Answer:
Kindly find the graphs attached
Explanation:
For figure 1: There is a steady increase in the position of the object as time increases. This is because despite the negative acceleration (deceleration), the object continues to move and cover more ground as time goes by.
<em>The straight line graph is observed because the acceleration is constant and not varying.</em>
For Figure 2: The graph of velocity vs time will have an inverted nature. This is because since the object is decelerating, it is reducing in its velocity as time goes by (increases). <em>This is also in a straight line since the deceleration is constant.</em>
Answer:
V = f λ speed of wave in terms of frequency and wavelength
t = S / V time for wave to travel a distance S
t = 91.4 m / 344.5 m/s = .265 sec time to travel 91.4 m
Answer:
O D.
Explanation:
Physics has an aspect that deals with the study of energy
