Answer:
The track's angular velocity is W2 = 4.15 in rpm
Explanation:
Momentum angular can be find
I = m*r^2
P = I*W
So to use the conservation
P1 + P2 = 0
I1*W1 + I2*W2 = 0
Solve to w2 to find the angular velocity
0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2
W2 = 0.435 rad/s
W2 = 4.15 rpm
Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is

where

is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by

where r is the radius of the wire. Substituting in the previous equation ,we find

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is

Therefore, the new resistivity must be 4 times the original one.
Let us consider two bodies having masses m and m' respectively.
Let they are separated by a distance of r from each other.
As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -
where G is the gravitational force constant.
From the above we see that F ∝ mm' and 
Let the orbital radius of planet A is
= r and mass of planet is
.
Let the mass of central star is m .
Hence the gravitational force for planet A is 
For planet B the orbital radius
and mass
Hence the gravitational force 
![f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }](https://tex.z-dn.net/?f=f_%7B2%7D%20%3DG%5Cfrac%7Bm%2A3m_%7B1%7D%20%7D%7B%5B2r_%7B1%7D%5D%20%5E%7B2%7D%20%7D)

Hence the ratio is 
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