Answer:
The potential difference between the places is 0.3 V.
∴ 1st option i.e. 0.3V is the correct option.
Explanation:
Given
Work done W = 3J
Amount of Charge q = 10C
To determine
We need to determine the potential difference V between the places.
The potential difference between the two points can be determined using the formula
Potential Difference (V) = Work Done (W) / Amount of Charge (q)
or
substituting W = 3 and q = 10 in the formula
V
Therefore, the potential difference between the places is 0.3 V.
∴ 1st option i.e. 0.3V is the correct option.
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
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Hey the answer to the question is
m = 0.40
