Answer:
Part a)

Part b)

Part c)

Part d)

Explanation:
Part a)
While bucket is falling downwards we have force equation of the bucket given as

for uniform cylinder we will have

so we have


now we have




now we have


Part b)
speed of the bucket can be found using kinematics
so we have



Part c)
now in order to find the time of fall we can use another equation



Part d)
as we know that cylinder is at rest and not moving downwards
so here we can use force balance



Answer:
A) 
B) F = 1632.65 N
Explanation:
Given details
outside air speed is given as 
since inside air is atmospheric , 
a) By using bernoulli equation between outside and inside of flight


![\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Crho%5B%20v_2%5E2%20-v_1%5E2%5D)
![\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%201.29%20%5B%20150%5E2%20-%200%5E2%5D)

b) force exerted on window
Area of window 
We know that force is given as


F = 1632.65 N
The altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg
<h3>How to determine the
altimeter reading?</h3>
The given parameters are:
- Ground level = 700 ft i.e. the field elevation
- Pressure altitude = 1450 ft
The formula to calculate the pressure altitude is:
Altitude = (29.92 – Altimeter reading) * 1,000 + Ground level
Substitute the known values
1450 = (29.92 - Altimeter reading) * 1000 + 700
Evaluate the like terms
750 = (29.92 - Altimeter reading) * 1000
Divide both sides by 1000
0.75 = 29.92 - Altimeter reading
Evaluate the like terms
Altimeter reading = 29.17
Hence, the altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg
Read more about pressure at:
brainly.com/question/26040104
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Answer:
The fastest satellite must change orbit
The most massive body (m₁) transfers more momentum to the satellite,
Explanation:
For this problem we consider a system formed by the satellite and each of the bodies with which it collides, in this system the forces during the collision are internal, the amount of movement must be conserved. Let's write the momentum is two instants
Most massive body (m1)
initial. Before the crash
p₀₁ = M v + m₁ v₁
after the crash
= M v´ + m₁ v₁´
how momentum is conserved
p₀ = p_{f}
Lighter body (m2)
p₀₂ = M v + m₂ v₂
p_{f2} = M v´ + m₂ v₂´
Let's clarify that the speed of the satellite and the object do not have the same direction, in general these shocks are elastic.
We can see that p₀₁> p₀₂
Let us analyze the two cases when the body collides, The most massive body (m₁) transfers more momentum to the satellite, therefore there must be a greater change in its momentum and velocity.
The fastest satellite must change orbit, thus rotating at a different distance from Earth