Answer:
option C
Explanation:
given,
diameter of circular room = 8 m
rotational velocity of the rider = 45 rev/min
= 
=4.712 rad/s
here in this case normal force is equal to centripetal force
N = m r ω²
N = m x 4 x 4.712²
N = 88.83m
frictional force = μ N
= 88.83m x μ
now, for the body to not to slide
gravity force is equal to frictional force
m g = 88.83 m x μ
g = 88.83 x μ
9.8 = 88.83 x μ
μ = 0.11
hence, the correct answer is option C
Answer:
C) 40 N/m
Explanation:
If we ASSUME that the spring is un-stretched at the zero cm position
k = F/Δx = 10/0.25 = 40 N/m
distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years
The acceleration due to gravity on Earth is 9.8 m/s per second.
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
