(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:
A. 
B. 
C. 
Explanation:
The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

is the capacitance,
is the common plate area,
is the plate separation and
is the permittivity of the material between the plates.
For air or free space,
is
called the permittivity of free space. In general,
where
is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum,
.
The energy stored in a capacitor is the average of the product of its charge and voltage.

Its charge,
, is related to its capacitance by
(this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for
,

A. Substituting for
in
,

B. When the distance is
,


C. When the distance is restored but with a dielectric material of dielectric constant,
, inserted, we have

Assuming that the students
worldwide are being considered, because of the extremely large population, this
can be considered as a binomial distribution. A normal distribution is used most
usually as a fair approximation of the binomial. The mean is the expectation,
therefore:<span>
E[x] = np = (16)(0.22) = 3.52
<span>μ = 3.52 </span></span>
I think iron? i’m not 100% sure