A weight lifter does 586 J of work on a weight that he lifts in 3.5 seconds. What is the power with which he lifts the weight?
1 answer:
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F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
Answer:
6.0 m below the top of the cliff
Explanation:
We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation
where
u = 0 (it starts from rest)
g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)
h = 24 m is the distance covered
Solving for h,
So the ball thrown upward is launched with this initial velocity:
u = 21.7 m/s
From now on, we take instead upward as positive direction.
The vertical position of the ball dropped from the cliff at time t is
While the vertical position of the ball thrown upward is
The two balls meet when
So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is
So the distance below the top of the cliff is