A rectangular tank of base 2m*1.5m and height 3m is filled with 7500kg of a liquid.The pressue at the middle of the tank is.
density of fluid = 7500 kg / 9 m^3 = 833 kg/m^3
1.5 meters of fluid above middle
so
p above one atmosphere = rho g h = 833 * 9.81 * 1.5 = 12,263 Pascals
To find the pressure at the middle of the tank, we can use the formula:
Pressure = weight / area
First, let's calculate the weight of the liquid in the tank:
Weight = mass × gravity
Where:
mass = 7500kg (given)
gravity = 9.8 m/s^2 (acceleration due to gravity)
Weight = 7500kg × 9.8 m/s^2
Weight = 73,500 N
Next, we need to find the area:
Area = length × width
Given:
Length = 2m
Width = 1.5m
Area = 2m × 1.5m
Area = 3m^2
Finally, we can substitute these values into the pressure formula:
Pressure = 73,500 N / 3m^2
Simplifying, we get:
Pressure = 24,500 N/m^2
So the pressure at the middle of the tank is 24,500 N/m^2.
To find the pressure at the middle of the tank, we can use the formula:
Pressure = Force / Area
First, let's calculate the force exerted by the liquid. The force is equal to the weight of the liquid:
Force = mass * gravitational acceleration
Given that the mass of the liquid is 7500 kg, and the gravitational acceleration is approximately 9.8 m/s², we can calculate the force:
Force = 7500 kg * 9.8 m/s² = 73,500 N
Now, we need to calculate the area of the base of the tank. The base of the tank is a rectangle with dimensions 2 m by 1.5 m, so the area is:
Area = length * width = 2 m * 1.5 m = 3 m²
Finally, we can calculate the pressure at the middle of the tank:
Pressure = Force / Area = 73,500 N / 3 m² = 24,500 Pa (Pascal)
Therefore, the pressure at the middle of the tank is 24,500 Pascal.