A. inelastic, since the girl moves in the same direction as the thrown ball
Answer:
m = 1 kg
Explanation:
Given that,
The force constant of the spring, k = 39.5 N/m
The frequency of oscillation, f = 1 Hz
The frequency of oscillation is given by the formula as formula as follows :
So, the mass that is attached to the spring is 1 kg.
Answer:
The magnitude of the change in momentum of the stone is 5.51kg*m/s.
Explanation:
the final kinetic energy = 1/2(0.15)v^2
1/2(0.15)v^2 = 70%*1/2(0.15)(20)^2
v^2 = 21/0.075
v^2 = 280
v = 16.73 m.s
if u is the initial speed and v is the final speed, then:
u = 20 m/s and v = - 16.73m/s
change in momentum = m(v-u)
= 0.15(- 16.73-20)
= -5.51 kg*m.s
Therefore, The magnitude of the change in momentum of the stone is 5.51kg*m/s.
Ok so this is a kinetic and potential energy problem. KE= 1/2*m*v^2 and PE= mgh so we set them equal to each other and get KE=PE so 1/2mv^2=mgh so mass divers out and left with 1/2v^2 =gh so solving for h or height we get
h=(1/2v^2)/g so h= 1/2*(6.5m/s)^2/9.80m/s^2 = 2.16m in height the ball will reach.
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