9. A heavy object was dropped from some distance
1 answer:
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Answer:
Minimum time interval (t2)=0.90 SECONDS
Explanation:
- coefficient of friction for employees footwear = 0.5
- coefficient of friction for typical athletic shoe = 0.810
- frictional force = coefficient of friction X acceleration due to gravity X mass of body
- Acceleration due to gravity is a constant = 9.81 m/s
- Let frictional force for employee footwear = FF1
- Let frictional force for athletic footwear =FF2
FF1 = O.5 X 9.81 X mass of body
= 4.905 x mass of body
FF2 = 0.810 X 9.81 X mass of body
= 7.9461 x mass of body
The body started from rest there by making the initial velocity zero ( u = 0)
From d= ut + 1/2 a x
- d =
x a x
.....................................i
where d= distance and it is given as 3.25m
- F =ma ...................................ii
making acceleration subject of the formula from equation ii
- a =
Making t subject of formula from equation (i)
- t=
where
-
= 4.905
=7.9461
Let
- t1 = minimum time taken for frictional force for employee foot wear
- t1 =
=1.15 seconds
- t2 =
= 0.90 seconds
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If total charge Q is enclosed in the surface of cube
then we will have say that total flux linked with all surfaces of the cube will be given by
since the position of charge is symmetric with respect to the center of the cube so here this whole flux will be equally linked with each of the face of the cube
So here total 6 faces of the cube is there and the total flux of the charge will equally divide in these 6 faces
so flux linked with each face is given by formula
so each face will have above flux due to central position of charge Q