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:
I think is D I'm sorry if I'm wrong
Answer:
The acceleration of the wagon is 3 m/s².
To calculate the acceleration of the wagon, we use the formula below.
Formula:
F = ma............. Equation 1
Where:
F = horizontal Force
m = mass of the wagon
a = acceleration of the wagon.
make a the subject of the equation
a = F/m.............. Equation 2
From the question,
Given:
F = 30 N
m = 10 kg
Substitute these values into equation 2
a = 30/10
a = 3 m/s²
Hence, the acceleration of the wagon is 3 m/s².
