A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is 2.5 m. (The acceleration due to gravity is 9.8 m/s2m/s2 and the density of water is 1000 kg/m3kg/m3.) How much work (in Joules) is required to:
1 answer:
Answer:
Explanation:
Diameter of pool = 12 m
radius of pool, r = 6 m
Total height raised, h = 3 + 2.5 = 5.5 m
density of water, d = 1000 kg/m³
Mass of water, m = Volume of water x density
m = πr²h x d
m = 3.14 x 6 x 6 x 5.5 x 1000
m = 113040 kg
Work = m x g x h
W = 113040 x 9.8 x 5.5
W = 6092856 J
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Answer:
centripetal acceleration
Explanation:
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97.8rpm = 97.8 * (π/30 rad/sec)
=10.25rad/sec
linear velocity= angular velocity *radius
radius =13.2cm=13.2/100=0.132m
v=rω
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