Answer:
The heat loss per unit length is 
Explanation:
From the question we are told that
The outer diameter of the pipe is 
The thickness is
The temperature of water is
The outside air temperature is 
The water side heat transfer coefficient is 
The heat transfer coefficient is 
The heat lost per unit length is mathematically represented as
![\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%5Cpi%20%28T%20-%20Ta%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_1%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_2%7D%7D)
Substituting values
![\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%2A%203.142%20%28363%20-%20263%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B300%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B20%7D%7D)


Answer:
White dwarfs are likely to be much more common. The number of stars decreases with increasing mass, and only the most massive stars are likely to complete their lives as black holes. There are many more stars of the masses appropriate for evolution to a white dwarf.
To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

Here,
P = Pressure
d = Diameter
t = Thickness
At the same time the longitudinal stress is given as,

The letters have the same meaning as before.
Then he hoop stress would be,



And the longitudinal stress would be



The Mohr's circle is attached in a image to find the maximum shear stress, which is given as



Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi
Answer:
The correct option is C
Explanation:
According to third equation of motion, v
2
=u
2
+2ax
Here, u=0 m/s
a=−g and x=−h
Negative sign indicates downward direction. Displacement and acceleration both are downwards.
So,v=±
2(−g)(−h)
We take minus sign because it is downwards.
v=−
2gh
After bouncing. velocity becomes 80% of v, i.e.,
v
′
=+0.8
2gh
(positive sign because the direction of ball has reversed after bouncing and is upwards.
Applying third equation of motion again, for u=v
′
, v=0 and a=−g
v
2
=u
2
+2×a×x
Thus,
0=0.64(2gh)+2(−g)x
or
x=0.64h