A rectangular tank is filled with ethyl alcohol to a full depth of 0.30m. The pressure at the surface of the alcohol is 1.013 x 105 Pa. Find the pressure at the bottom of the tank. The density of ethyl alcohol is 789 kg/m3.
We can use the formula for pressure at a depth in a liquid:
P = ρgh
Where:
P = pressure
ρ = density of the liquid
g = acceleration due to gravity
h = depth of the liquid
At the surface of the alcohol, h = 0, so we can use this to find g:
P = ρgh
1.013 x 105 = (789 kg/m3)g(0)
g = 0
This gives us g = 0, which doesn't seem right. But the reason for this is that the pressure at the surface of the liquid is atmospheric pressure, which already takes into account the weight of the liquid above it. So the pressure contribution from the weight of the liquid is zero at the surface.
Now we can find the pressure at the bottom of the tank:
P = ρgh
P = (789 kg/m3)(9.81 m/s2)(0.30 m)
P = 2325.87 Pa
Therefore, the pressure at the bottom of the tank is 2325.87 Pa.
What volume (in cc) will 400g of silver occupy if its density is 10.5 g/ml?
We can use the formula for density:
density = mass/volume
Rearranging this formula, we get:
volume = mass/density
Substituting the given values, we get:
volume = 400g / 10.5 g/ml
Simplifying this, we get:
volume = 38.1 ml
Since 1 ml is equal to 1 cc, the volume of 400g of silver is 38.1 cc.
Atmospheric pressure is about 1.0 x 105 Pa. How large a force does the still air in the room exert on the inside of a window pane that 60cm x 120cm?
The force exerted by the still air in the room on the inside of the window pane is equal to the atmospheric pressure multiplied by the area of the window pane.
The area of the window pane is:
A = (60 cm) x (120 cm) = 7200 cm²
Converting to square meters, we get:
A = 0.72 m x 1.2 m = 0.864 m²
The force exerted by the still air is:
F = P x A
where P is the atmospheric pressure, which is 1.0 x 10^5 Pa.
Substituting the values, we get:
F = (1.0 x 10^5 Pa) x (0.864 m²) = 86,400 N or 86.4 kN
Therefore, the still air in the room exerts a force of 86.4 kN on the inside of the window pane.
A 100-kg metal cylinder, 2.0m long and with each end of 25 cm^2, stands vertically on one end. What pressure that the cylinder exerts on the floor?
The pressure that the cylinder exerts on the floor is equal to the weight of the cylinder divided by the area of the end that is in contact with the floor.
The weight of the cylinder is:
W = m * g = 100 kg * 9.81 m/s^2 = 981 N
The area of one end of the cylinder is:
A = 25 cm^2 = 0.0025 m^2
Since the cylinder is standing vertically on one end, only one end is in contact with the floor, so we will use this area to calculate the pressure.
The pressure that the cylinder exerts on the floor is:
P = W / A = 981 N / 0.0025 m^2 = 392400 Pa
Therefore, the cylinder exerts a pressure of 392400 Pa on the floor.
A 100-kg metal cylinder, 2.0m long and with each end of 25 cm^2, stands vertically on one end. What pressure that the cylinder exerts on the floor?
The pressure that the cylinder exerts on the floor is equal to the weight of the cylinder divided by the area of the end that is in contact with the floor.
The weight of the cylinder is:
W = m * g = 100 kg * 9.81 m/s^2 = 981 N
The area of one end of the cylinder is:
A = 25 cm^2 = 0.0025 m^2
Since the cylinder is standing vertically on one end, only one end is in contact with the floor, so we will use this area to calculate the pressure.
The pressure that the cylinder exerts on the floor is:
P = W / A = 981 N / 0.0025 m^2 = 392400 Pa
Therefore, the cylinder exerts a pressure of 392400 Pa on the floor.
To find the pressure at the bottom of the tank, we can use the concept of hydrostatic pressure. The hydrostatic pressure at a certain depth within a fluid is given by the equation:
P = ρgh
Where:
P is the pressure,
ρ (rho) is the density of the fluid,
g is the acceleration due to gravity, and
h is the depth of the fluid.
First, we need to find the depth of the fluid in the tank. Since the tank is rectangular, we assume that the depth is equal to the height of the tank.
Given:
Height of the tank = 0.30 m
Density of ethyl alcohol (ρ) = 789 kg/m³
Now, we need to find the gravitational acceleration (g) to use it in the equation. The value of acceleration due to gravity is approximately 9.8 m/s².
Using the equation P = ρgh, we can calculate the pressure at the bottom of the tank.
P = 789 kg/m³ * 9.8 m/s² * 0.30 m
P = 2310.9 N/m² or Pascal (Pa)
Therefore, the pressure at the bottom of the tank is approximately 2310.9 Pa.