Answer:
Where 
represent the force for each of the 5 cases 
presented on the figure attached.
Explanation:
For this case the figure attached shows the illustration for the problem
We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.
Th formula is given by:
Where G is a constant 
represent the mass for the earth
represent the mass for the spaceship
represent the radius between the earth and the spaceship
For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.
Based on this case we can create the following rank:
Where represent the force for each of the 5 cases presented on the figure attached.
The strength of the fireman in vertical direction will be given by F = m * g. Then, the work done will be given by definition by W = F * d. Substituting the expression of the Force in that of the work, we have that the work will be W = m * g * d. Substituting the given values and assuming that g = 10m / s ^ 2, we have a total work of W = (73) * (10) * (9) = 6570 J
Answer:
Torque = R X F = R * F sin theta
-f * 2R will exert an equal opposite torque
-f * 2 = F sin theta
f = -F sin theta / 2
Answer:
Thus, the time for the first lamp is 44 minutes.
Explanation:
Power of first lamp, P' = 1000 W
Power of second lamp, P'' = 4400 W
time for second lamp, t'' = 10 minutes
Let the time for first lamp is t'.
As the energy is same, so,
P' x t' = P'' x t''
1000 x t' = 4400 x 10
t' = 44 minutes
Answer:
a) that laser 1 has the first interference closer to the central maximum
c) Δy = 0.64 m
Explanation:
The interference phenomenon is described by the expression
d sin θ = m λ
Where d is the separation of the slits, λ the wavelength and m an integer that indicates the order of interference
For the separation of the lines we use trigonometry
tan θ = sin θ / cos θ = y / x
In interference experiments the angle is very small
tan θ = sin θ = y / x
d y / x = m λ
a) and b) We apply the equation to the first laser
λ = d / 20
d y / x = m d / 20
y = m x / 20
y = 1 4.80 / 20
y = 0.24 m
The second laser
λ = d / 15
d y / x = m d / 15
y = m x / 15
y = 0.32 m
We can see that laser 1 has the first interference closer to the central maximum
c) laser 1
They ask us for the second maximum m = 2
y₂ = 2 4.8 / 20
y₂ = 0.48 m
For laser 2 they ask us for the third minimum m = 3
In this case to have a minimum we must add half wavelength
y₃ = (m + ½) x / 15
m = 3
y₃ = (3 + ½) 4.8 / 15
y₃ = 1.12 m
Δy = 1.12 - 0.48
Δy = 0.64 m
