Work = force x distance
= 100N (force) x 0.5m (distance)
= 50J
The direction of the electric field would be south.
qE/m = 115
<span> E = 115*m/q </span>
<span> = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) </span>
<span> = 762.87 * 10^(-12) </span>
<span> = 6.27 x 10^-10 N/C
</span>
Hope this answers the question. Have a nice day. Feel free to ask more questions.
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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