An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits
1 answer:
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Answer with Explanation:
We are given that
Distance,r=0.27 m
Tangential speed=v=0.49 m/s
a.Angular speed ,
Using the formula
Time period,
b.Amplitude,A=Distance of small eraser from the center of a turnable =0.27 m
c.Maximum speed,
d.Maximum acceleration=
Explanation:
The initial kinetic energy for both blocks is zero. Let
and
. So using the conservation law of linear momentum, we can write
or
The final kinetic energies for the two masses are
Therefore, the ratio of their kinetic energies is
Answer:
26.6 cm
Explanation:
We are given that
Spring constant=23.6 N/m
Force exert=6.28 N
We have to find the displacement.
We know that Hooke's law
Where k= Spring constant
x=Displacement
Using the formula
Then, we get
We know that 1 m=100 cm
Hence, the displacement =26.6 cm