(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:
The sound intensity level in the car is 57.2 dB.
Explanation:
Sound intensity level in decibels, β = 10 log (I/I₀); where I = 0.525 × 10⁻⁶ W/m², I₀ = 1.0 × 10⁻¹² W/m²
β (dB) = 10 log ((0.525 × 10⁻⁶)/(1.0 × 10⁻¹²)) = 10 × 5.72 = 57.2 dB
Hope this Helps!!!
Answer:
Explanation:
It would actually be A. 30 , as each hour of ascension (i am not sure about the correct terminology) equals 15 .
The car travels at a speed of 25m/s.
<u>Explanation:</u>
Given-
Mass, m = 1500kg
Coefficient of friction, μk = 0.47
Distance, x = 68m
Speed, s = ?
We know,
and
F = μ X m X g
Therefore,
μ * m * g = m * a
μ * g = a
Let, g = 9.8m/s²
So,
We know,
where, v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
If the car comes to rest, the final velocity, v becomes 0.
So,
The car travels at a speed of 25m/s.
