(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:
8.854 pF
Explanation:
side of plate = 0.1 m ,
d = 1 cm = 0.01 m,
V = 5 kV = 5000 V
V' = 1 kV = 1000 V
Let K be the dielectric constant.
So, V' = V / K
K = V / V' = 5000 / 1000 = 5
C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F
C = 8.854 pF
Explanation:
Optics is a branch of physics that is the study of light and vision. ... The branch of physics dealing with the nature and properties of electromagnetic energy in the light spectrum and the phenomena of vision. In the broadest sense, optics deals with infrared light, visible light, and ultraviolet light.
Answer:
Work Done = 67.5 J
Explanation:
First we find the value of spring constant (k) using Hooke's Law. Hooke's is formulated as:
F = kx
where,
F = Force Applied = 450 N
k = Spring Constant = ?
x = Stretched Length = 30 cm = 0.3 m
Therefore,
450 N = k(0.3 m)
k = 450 N/0.3 m
k = 1500 N/m
Now, the formula for the work done in stretching the spring is given as:
W = (1/2)kx²
Where,
W = Work done = ?
k = 1500 N/m
x = 70 cm - 40 cm = 0.3 m
Therefore,
W = (1/2)(1500 N/m)(0.3 m)²
<u>W = 67.5 J</u>
Answer:
I = 97.2 10³⁶ kg m²
Explanation:
The moment of inertia of a body the expression of inertia in the rotational movement and is described by the expression
I = ∫ r² dm
In this problem we are told to use the moment of inertia of a uniform sphere, the expression of this moment of inertia is
I = 2/5 M r²
where m is the mass of the earth and r is the radius of the earth.
Let's calculate
I = 2/5 5.97 10²⁴ (6.38 10⁶)²
I = 97.2 10³⁶ kg m²
