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
Answer:
It will have aim at a point "below" the insect.
From the insect's point of view, the fish will appear to be shallower than it actually is because a ray of light from the insect to the fish will be bent "towards" the normal when the ray enters the water
Answer:
open cluster. globular cluster. eclipsing binary
Solution :
Frequency may be defined as the number of observation or number of waves that is taken in per unit time. The unit of frequency is Hertz or Hz.
It is given that :
Successive harmonic frequencies, f = 52.2 Hz
and f' = 60.9 Hz
Therefore, fundamental frequency, F = f' - f
F = 60.9 - 52.2
F = 8.7 Hz
Therefore the string which is fixed at both the ends forms all the harmonics.
