Just like mass, energy, linear momentum, and electric charge, angular momentum is also conserved.
The wheel has angular momentum. I don't remember whether it's
up or down (right-hand or left-hand rule), but it's consistent with
counterclockwise rotation as viewed from above.
When you grab the wheel and stop it from spinning (relative to you),
that angular momentum has to go somewhere.
As I see it, the angular momentum transfers through you as a temporary
axis of rotation, and eventually to the merry-go-round. Finally, all the mass
of (merry-go-round) + (you) + (wheel) is rotating around the big common
axis, counterclockwise as viewed from above, and with the magnitude
that was originally all concentrated in the wheel.
(a) The stress in the post is 1,568,000 N/m²
(b) The strain in the post is 7.61 x 10⁻⁶
(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.
<h3>Area of the steel post</h3>
A = πd²/4
where;
d is the diameter
A = π(0.25²)/4 = 0.05 m²
<h3>Stress on the steel post</h3>
σ = F/A
σ = mg/A
where;
- m is mass supported by the steel
- g is acceleration due to gravity
- A is the area of the steel post
σ = (8000 x 9.8)/(0.05)
σ = 1,568,000 N/m²
<h3>Strain of the post</h3>
E = stress / strain
where;
- E is Young's modulus of steel = 206 Gpa
strain = stress/E
strain = (1,568,000) / (206 x 10⁹)
strain = 7.61 x 10⁻⁶
<h3>Change in length of the steel post</h3>
strain = ΔL/L
where;
- ΔL is change in length
- L is original length
ΔL = 7.61 x 10⁻⁶ x 2.5
ΔL = 1.9 x 10⁻⁵ m
Learn more about Young's modulus of steel here: brainly.com/question/14772333
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Answer:
Explanation:
Given
Total time=27 min 43.6 s=1663.6 s
total distance=10 km
Initial distance 
time taken=25 min =1500 s
initial speed 
after 8.13 km mark steve started to accelerate
speed after 60 s
distance traveled in 60 sec
time taken in last part of journey
distance traveled in this time
and total distance
