Hi there!
We know that:
U (Potential energy) = mgh
We are given the potential energy, so we can rearrange to solve for h (height):
U/mg = h
g = 9.81 m/s²
m = 30 g ⇒ 0.03 kg
0.062/(0.03 · 9.81) = 0.211 m
Given:
Lens.........diameter ...fl#
eyepiece...2cm............5
objective...40cm........15
focal length of eyepiece = 2*5 = 10cm
focal length of objective = 40*15 = 600cm
magnification = FL obj / FL eyp = 600/10 = 60x
Answer:
Work done, W = 6 J
Explanation:
It is given that,
Force of gravity acting on the book, weight of the book is 15 N
We need to find the work done in lifting the book straight up for a distance of 0.4 meters.
The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

So, the magnitude of work done in lifting the book is 6 joules.
Cubic centimeters for the volume of a solid.
Liters for volume of a liquid.
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v
= ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v
= ( u + u ) / ( 1 + ( u×u / c²) )
v
= 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v
= 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v
= 1.904c / ( 1 + 0.906304 )
v
= 1.904c / 1.906304
v
= 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c