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
Now
- Lower mass=Higher acceleration
- Lower Force=Lower Acceleration
Option B has lowest mass and highest force hence its correct
Answer:
at the top of the tree I hope it will help you please follow me
<span>The
_______ is the the distance between two crests or two troughs on a
transverse wave. It is also the distance between compressions or the
distance between rarefactions on a longitudinal wave.</span>
Answer:
option C
Explanation:
Let mass of the bullet be m and velocity be v
mass of gun be M and bullet be V
now,
using conservation of momentum for gun 1
(M+m) V' = 2 mv + 3 MV
V' = 0
3 M V = - 2 mv
momentum of gun 1 =- 2 mv---------(1)
now for gun 2
(M+m) V' = mv + MV
V' = 0
M V = - mv
momentum of gun 1 = -mv-----------(2)
dividing equation (1) by (2)
the correct answer is option C
