Problem: Suppose that the rate of flow of water through a fire hose is 21.4 kg/s and the stream of water from the hose moves at
2 answers:
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Answer:
<em>a) 105935.7 Pa</em>
<em>b) 103630.35 Pa</em>
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 =
where
V is the volume of the cylinder
h is the height of the cylinder
substituting values, we have
0.025 = 3.142 x 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 = <em>105935.7 Pa</em>
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 = <em>103630.35 Pa</em>
Explanation:
Given
initial velocity(v_0)=1.72 m/s
using
Where v=final velocity (Here v=0)
u=initial velocity(1.72 m/s)
a=acceleration
s=distance traveled
s=0.214 m
(b)time taken to travel 0.214 m
v=u+at
t=0.249 s
(c)Speed of the block at bottom
Here u=0 as it started coming downward
v=1.72 m/s