Answer:
Explanation:
Potential energy on the surface of the earth
= - GMm/ R
Potential at height h
= - GMm/ (R+h)
Potential difference
= GMm/ R - GMm/ (R+h)
= GMm ( 1/R - 1/ R+h )
= GMmh / R (R +h)
This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.
Extra energy is needed to get the same object into orbit at height h
= Kinetic energy of the orbiting object at height h
= 1/2 x potential energy at height h
= 1/2 x GMm / ( R + h)
Let t=time to reach the ground=8 secs, g= acceleration of gravity. The speed v on reaching the ground is gt=8g=78.4 m/s where g=9.8 m/s/s approx.
Answer:
W = 2352 J
Explanation:
Given that:
- mass of the bucket, M = 10 kg
- velocity of pulling the bucket, v = 3

- height of the platform, h = 30 m
- rate of loss of water-mass, m =

Here, according to the given situation the bucket moves at the rate,

The mass varies with the time as,

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance
∆x = 3∆t meters
So, during this interval change in work done,
∆W = m.g∆x
<u>For work calculation:</u>
![W=\int_{0}^{10} [(10-0.4t).g\times 3] dt](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7B10%7D%20%5B%2810-0.4t%29.g%5Ctimes%203%5D%20dt)
![W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}](https://tex.z-dn.net/?f=W%3D%203%5Ctimes%209.8%5Ctimes%20%5B10t-%5Cfrac%7B0.4t%5E%7B2%7D%7D%7B2%7D%5D%5E%7B10%7D_%7B0%7D)

Answer:
The velocity of water at the bottom, 
Given:
Height of water in the tank, h = 12.8 m
Gauge pressure of water, 
Solution:
Now,
Atmospheric pressue, 
At the top, the absolute pressure, 
Now, the pressure at the bottom will be equal to the atmopheric pressure, 
The velocity at the top,
, l;et the bottom velocity, be
.
Now, by Bernoulli's eqn:

where

Density of sea water, 


