(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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As momentum / time = force
so; time = 100÷15
so your answer is 6.7 !!
The truck has more KE than the bike
We use the equation of motion for vertical component,
Here, 
is displacement of bullet, 
is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.
Therefore,
Given, 
Substituting the values, we get time of flight
work done = force * distance moved (in direction of the force)
force= mass* acceleration
force=58.1N
58.1*(5.8*10^4)
=3,369,800 J
