Answer:
(a) 625 ft lb
(b) 468.75 ft lb
Explanation:
Work=Force* Distance
(a)
Force=wl/2 where w is uniformly distributed load of 0.5 lb/ft and l is the length of the rope
Force=0.5*50/2=12.5
Work=12.5*50=625 ft lb
(b)
Assuming that the weight of the rope is concentrated in two places, half is 12.5 feet down and the other half is 37.5 feet down. The center of mass of the top half is 12.5 feet down. The weight of the top half is 12.5
Work=work on top half + work on bottom half
Work on top half=12.5*(0.5*25)=156.25 ft lb
work on bottom half=12.5*(0.5*50)=312.5 ft lb
Total work=156.25+312.5=468.75 ft lb
Answer:
h = 0.2 m
Explanation:
Given that,
A pendulum is moving 2.0 m/s at the bottom of its swing.
We need to find the height high it swing back. Let the height is h.
Using the conservation of energy such that,

Put all the values,

So, it will reach to a height of 0.2 m.
Newton's second law is stated as:
F=ma,
a = (7-4)/1.5 = 2 m/s^2 (it is a deceleration due to impact with floor. And thus the ball exerts force on the floor).
Therefore,
F= 0.3*2 = 0.6 N
This question is based on the simple principle of the refraction.
Refraction is the optical phenomenon in which the light ray will change its direction when fall obliquely on a refracting surface due to the change of velocity of light in the refracting medium.
Greater the refractive index,the lesser will be the velocity of light in that medium.
As refractive index
Here
is the velocity of light in first medium and
is the velocity of light in second medium.
In case of light travelling from space to the medium we may write
where c is the velocity of light.
So as long as the medium is same ,there will be no change in the velocity of light.
As per the question light was initially in space. After then the light entered into glass medium whose refractive index is more as compared to space.Hence velocity of light will be decreased .
Now the light escaped through the glass.so it again enters the same medium i.e space. Hence the velocity of light will be increased now and will be equal to 'c' now .