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Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
Explanation:
(1). Formula to calculate the potential difference is as follows.
= 
= 
= 
= 
= 38.7 volts
Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.
(2). Now, formula to calculate the energy stored in the capacitor is as follows.
E = 
= 
= 
Thus, the electric-field energy stored in the capacitor is .
<h2>
Displacement along west = 3612 m</h2><h2>
Displacement along north = 4633.50 m</h2>
Explanation:
Let east be positive x axis and north be positive y axis
Velocity of boat = 2.8 m/s in a direction 52° north of west.
Velocity, v = -2.8 cos 52 i + 2.8 sin 52 j = -1.72 i + 2.21 j m/s
Time taken = 35 min = 35 x 60 = 2100 s
Displacement = Velocity x Time
Displacement = (-1.72 i + 2.21 j) x 2100
Displacement = -3612 i + 4633.50 j m
Displacement along west = 3612 m
Displacement along north = 4633.50 m
