On Earth, none of the atmosphere is.
Answer:
Yes, you would consider the mass × gravity at the centre of gravity of the object in question to cause a moment, hence it should be included in your resolution
The moment of inertia is 
Explanation:
The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.
The moment of inertia of the rod about its centre is given by

where
M = 24 kg is the mass of the rod
L = 0.96 m is the length of the rod
Substituting,

The moment of inertia of one ball is given by

where
m = 50 kg is the mass of the ball
is the distance of each ball from the axis of rotation
So we have

Therefore, the total moment of inertia of the system is

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Answer:
Explanation:
Part A) Using
light intensity I= P/A
A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2
Radius= Diameter/2
P= power= 10*10^-3=0.01 W
light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2
Part B) Using
I=c*ε*E^2/2
rearrange to solve for E=
((I*2)/(c*ε))
c is the speed of light which is 3*10^8 m/s^2
ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1
I= the already solved light intensity= 8.85*10^10 W/m^2
amplitude of the electric field E=
(9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)
---> E=
(1.8*10^11) / (2.66*10^-3) =
(6.8*10^13) = 8.25*10^6 V/m
Answer:
The work done on the suitcase is, W = 1691 J
Explanation:
Given data,
The force on the suitcase is, F = 89 N
The distance Russell dragged the suitcase, S = 19 m
The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction
The work done on the suitcase by Russell is given by the formula
W = F · S
Substituting the given values,
W = 89 N x 19 m
W = 1691 J
Hence, the work done on the suitcase is, W = 1691 J