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Answer:
39.2m/s
Explanation:
The potential energy the book has right before it falls is equal to the kinetic energy in falling.
PE = KE
mgh = (1/2)mv
2gh=v
v=(2)(9.81)(2)
v=39.24m/s
Answer:
1) t=1.743 sec
2)Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)Vf=17.08 m/s
Explanation:
1)From second equation of motion we get
h=Vit+(1/2)gt^2
here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion
14.9=(0)*t+(1/2)(9.8)t^2
t^2=14.9/4.9
t^2=3.040 sec
t=1.743 sec
2) s=Vo*t
Putting values we get
107=Vo*1.743
Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)From third equation of motion we know that
Vf^2-Vi^2=2gh
here Vi=0 m/s,h=14.9 m
Vf^2=Vi^2+2gh=0+2(9.8)(14.9)
Vf^2=292.04
Vf=17.08 m/s
Answer:
v=0.60 m/s
Explanation:
Given that
m ₁= 390 kg ,u ₁= 0.5 m/s
m₂ = 250 kg ,u₂ = 0.76 m/s
As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.
Pi = Pf
m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v
Now putting the values in the above equation
390 x 0.5 + 250 x 0.76 = (390 + 250 ) v
v=0.60 m/s
Therefore the velocity of the system will be 0.6 m/s.
