(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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As per the question the mass of the boy is 50 kg.
The boy sits on a chair.
We are asked to calculate the force exerted by the boy on the chair at sea level.
The force exerted by boy on the chair while sitting on it is nothing else except the force of gravity of earth i.e the weight of the body .The direction of that force is vertically downward.
At sea level the acceleration due to gravity g = 9.8 m/s^2
Hence the weight of the boy
[m is the mass of the body]
we have m = 50 kg.
Hence w = 50 kg ×9.8 m/s^2
=490 N kg m/s^2
= 490 N
Here newton [N] is the unit of force.
Answer:
The kinetic energy of the ejected electrons increases.
Explanation:
As we know that electrons are only ejected from a metal surface if the frequency of the incident light increases the work function of the metal. If the frequency of the incident light is less than the work function of the metal no matter how intense the beam the electrons will not be ejected from the surface.
Using conservation of energy principle we have
If we increase the intensity of incident light the term on the LHS of the above equation increases this increase appears in the kinetic energy term in RHS of the equation since
remains constant.
For the part a) we need only the momentum of the box and we have the data to find it.
Momentum is given by,
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where clearly, p is the momentum, m the mass of the box and v is the velocity.
Substituting,

For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,

<em>It is the same that part a)</em>