The normal force is the supporting force that is exerted on an object that is in contact with another stable object.
Answer: Option C
<u>Explanation:
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Normal force is forward or upward pushing force acting on an object. Mostly the normal force acts as supporting force exerted on the object by the neighbouring stable object with which the object in question is in contact. So normal force falls under the category of contact forces.
Generally, normal force will be acting to support the weight of any object placed on another object. The best examples of normal forces are the weight of the book supported by table or by the pushing force of the wall on the person leaning on the wall.
Answer:
C
Explanation:
The kicker exerts more force of the football which is why the football moves when its kicked.
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
