(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 is B.
In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.
Hope it helped you.
-Charlie
Answer:
<em>J=36221 Kg.m/s</em>
Explanation:
<u>Impulse-Momentum Theorem</u>
These two magnitudes are related in the following way. Suppose an object is moving at a certain speed 
and changes it to 
. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by
The initial and final momentums are, respectively
The change of momentum is
It is numerically equal to the Impulse J
We are given
The impulse the car experiences during that time is
J=-36221 Kg.m/s
The magnitude of J is
J=36221 Kg.m/s
