The work required is Wa = 2954112 J
Given:
swimming pool diameter = 14 m
length of sides = 4 m
height of water = 3 m
To Find:
work required to pump water
Solution: The radius of the swimming pool is
r = 14/2 = 7 m
The work is mathematically given as
W = Force x distance
Now force is mathematically given as
F = density x area x height of pool = p*(πr²)dx
Now the work done to pump all of the water over the side
W = ∫p*(πr²)(H-x)dx = ∫1000*9.81*(π*7^2)(4-x)dx
W = 64000*9.8π∫(4-x) dx = 64000*9.8π{4(3) - 3/2}
W = 2954112 J
So, work required is Wa = 2954112 J
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Simile, as the example compares the person with a cheetah using “as” (a simile uses “like” or “as” for comparisons between two things).
Answer: 7 - Transpiration, 6 - Infiltration, 5 - Runoff, 4 - Evaporation, 3 - Precipitation, 2 - Condensation, 1 - Sublimation,
Explanation:
1.0 joule= 1.0 newtons × 1.0 meter = 1.0 newton × meter
Work = 10 newtons × 5 meters = 50 newton × meter
Answer:
60 rad/s
Explanation:
∑τ = Iα
Fr = Iα
For a solid disc, I = ½ mr².
Fr = ½ mr² α
α = 2F / (mr)
α = 2 (20 N) / (0.25 kg × 0.30 m)
α = 533.33 rad/s²
The arc length is 1 m, so the angle is:
s = rθ
1 m = 0.30 m θ
θ = 3.33 rad
Use constant acceleration equation to find ω.
ω² = ω₀² + 2αΔθ
ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)
ω = 59.6 rad/s
Rounding to one significant figure, the angular velocity is 60 rad/s.
