Answer:
a) 105935.7 Pa
b) 103630.35 Pa
Explanation:
The volume of the container = 0.025 m^3
The radius of the container = 13 cm = 0.13 m
We have to find the height of the tank
From the equation for finding the volume of the cylinder,
V = [tex]\pi r^2h[/tex]
where
V is the volume of the cylinder
h is the height of the cylinder
substituting values, we have
0.025 = 3.142 x [tex]0.13^2[/tex] x h
0.025 = 0.0531h
h = 0.025/0.0531 = 0.47 m
Pressure at the bottom of the tank P = ρgh
where
ρ is the density of water = 1000 kg/m^3
g is the acceleration due to gravity = 9.81 m/s^2
h is the depth of water which is equal to the height of the tank
substituting values, we have
P = 1000 x 9.81 x 0.47 = 4610.7 Pa
atmospheric pressure = 101325 Pa
therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = 105935.7 Pa
b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m
pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa
This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = 103630.35 Pa
(a) The pressure exerted by water on the sides of the container at the bottom is 4606 Pa.
(b) The pressure exerted by water on the sides of the container halfway down from the top is of 2303 Pa.
Given data:
The volume of cylindrical container is, [tex]V = 0.025 \;\rm m^{3}[/tex].
The radius of container is, r = 13 cm = 0.13 m.
(a)
When the container is full, the pressure that the water exerts on the sides of the container is given as,
[tex]P = \rho gh[/tex]
Here, [tex]\rho[/tex] is the density of water, g is the gravitational acceleration and h is the height of container and its value is obtained as,
[tex]V = \pi r^{2}h\\\\0.025= \pi \times (0.13)^{2} \times h\\\\h = 0.47 \;\rm m[/tex]
Then pressure is,
[tex]P = 1000 \times 9.8 \times 0.47\\\\P=4606 \;\rm Pa[/tex]
Thus, the pressure exerted by water on the sides of the container at the bottom is 4606 Pa.
(b)
For the ides of the container halfway down from the top, the height is,
h' = h/2
h' = 0.47/2 = 0.235 m
Then the pressure is obtained as,
[tex]P' = \rho \times g \times h'\\\\P' = 1000 \times 9.8 \times 0.235\\\\P'=2303 \;\rm Pa[/tex]
Thus, the pressure exerted by water on the sides of the container halfway down from the top is of 2303 Pa.
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