(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:
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Explanation:
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Answer:
25.33 rpm
Explanation:
M = 100 kg
m1 = 22 kg
m2 = 28 kg
m3 = 33 kg
r = 1.60 m
f = 20 rpm
Let the new angular speed in rpm is f'.
According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.
Initial angular momentum = final angular momentum
(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =
(1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'
(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'
( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'
2660 = 105 x f'
f' = 25.33 rpm
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