P = m x v
P = 30 x 10
=300
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
Energy is "the ability to do work". Energy is how things change and move. It takes energy to cook food, to drive to school, and to jump in the air. Different forms of Energy. Energy can take a number of different forms.
Answer:
refraction
Explanation: hope this helps :)
Answer:-q
Explanation:
Given
Capacitor is charged to a battery and capacitor acquired a charge of q i.e.
+q on Positive Plate and -q on negative Plate.
If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by
where A=area of capacitor plate
d=Separation between plates
This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q
