The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
Answer:
no where is the main part of the question dude
Answer:
Explanation:
Work done on the lever ( input energy ) = force applied x input distance
= 24 N x 2m = 48 J
Work done by the lever ( output energy ) = load x output distance
= 72 N x 0.5m = 36 J
efficiency = output energy / input energy
= 36 J / 48 J
= 3 / 4 = .75
In percentage terms efficiency = 75 % .
Answer:
199.0521 Will be the answer
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
