If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.
Answer:
Explanation:
Impulse = change in momentum
mv - mu , v and u are final and initial velocity during impact at surface
For downward motion of baseball
v² = u² + 2gh₁
= 2 x 9.8 x 2.25
v = 6.64 m / s
It becomes initial velocity during impact .
For body going upwards
v² = u² - 2gh₂
u² = 2 x 9.8 x 1.38
u = 5.2 m / s
This becomes final velocity after impact
change in momentum
m ( final velocity - initial velocity )
.49 ( 5.2 - 6.64 )
= .7056 N.s.
Impulse by floor in upward direction
= .7056 N.s
Answer:
B. τ = 16 Nm
Explanation:
In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:
τ = Fd
here,
τ = Torque = ?
F = Force exerted by the weight = Weight = mg
F = mg = (4 kg)(10 m/s²) = 40 N
d = distance from knee to weight = 40 cm = 0.4 m
Therefore,
τ = (40 N)(0.4 m)
<u>B. τ = 16 Nm</u>
