The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.
Given that, the tank is 8 feet across the top and 6 feet high
By the property of similar triangles, 4/6 = r/y
6r = 4y
r = 4/6*y = 2/3*y
Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²
The weight of each disc is m = ρw* A
m = 62.4* 4π/9*y² = 87.08*y²
The distance pumped is 6-y.
The work done in pumping the tank by pumping the water over the top edge is
W = 87.08 ∫(6-y)y² dy
W = 87.08 ∫(6y³ - y²) dy
W = 87.08 [6y⁴/4 - y³/3]
W = 87.08 [3y⁴/2- y³/3]
The limits are from 0 to 6.
W = 87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs
The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.
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Answer:
The work done required on the coin during the displacement is 21.75 w.
Explanation:
Given that,
A coin slides over a friction-less plane i.e friction force = 0.
The co-ordinate of the given point is (1.40 m, 7.20 m).
The position vector of the given point is represented by
.
The displacement of the coin is

The force has magnitude 4.50 N and its makes an angle 128° with positive x axis.
Then x component of the force = 4.50 cos128°
The y component of the force = 4.50 sin128°
Then the position vector of the force is


We know that,
work done is a scalar product of force and displacement.


=(-2.77×1.40+ 3.56×7.20) w
=21.75 w
The work done required on the coin during the displacement is 21.75 w.
Answer:
16.63265306 Kg
Explanation:
This is a Newtons to weight conversion.
Fg = weight
Fg = N/9.8