Answer:
Work done in all the three cases will be the same.
Explanation:
1) The free falling body has only one force acting on it, the gravitational force. The work done on the body = mgH (Gravitational potential energy)
2) There are two forces acting on the body going down on a frictionless inclined plane - gravity and the normal force. The gravitational potential energy will be the same. The work done due to the normal force is zero, since the direction of the force is perpendicular to the displacement. Hence, total work done on the body = mgH
3) In the case of the body swinging on the end of a string, the change in gravitational potential enrgy will once again be the same since difference in height is H. The additional force on the body is the tension due to the string. But the work done due to this force is <em>zero, </em>since the displacement of the body is perpendicular to the tension. Therefore, the total work done on the body is once again mgH.
Answer:
and
and
Explanation:
= Wavelength
= Angle
m = Order
Distance between grating is given by
We have the relation
m = 1
m = 2
The first and second order angular deflection is and
m = 1
m = 2
The first and second order angular deflection is and .
You should use math on your mind and you should balance the bottle
Answer:
45.5 m
Explanation:
m = 2 kg, h = 20 m, E = 500 J, radius of earth = R, mass of earth = M
find the new height H
at h, the potential energy = -GMm/(R + h)
at H, the potential energy = -GMm/(R + H)
increase of the potential energy
= [-GMm/(R + H)] - [-GMm(R + h)]
= GMm[1/(R + h) - 1/(R + H)] = E
1/(R + h) - 1/(R + H) = E/(GMm)
(H - h)/[(R + H)(R + h)] = E/(GMm)
R + h ≈ R, R + H ≈ R
so (H - h)/R² = E/(GMm)
H - h = ER²/(GMm)
note GM/R² = g = 9.81 m/s²
so H - h = E/(mg)
H = h + E/(mg) = 20 + 500/(2*9.81) = 45.5 m
Answer:
The net force is 21 N, and the bag will go to the right.
Explanation:
You can find the net force by adding both values of the boy and the girl, and then subtracting four from it. You can also figure out which direction its moving by looking at which side is exerting the most force, which would be the right side.
<em>Hope </em><em>it </em><em>helps</em><em>!</em>