(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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The 'strength' of the electric field is the force on 1C of charge at that point.
At this 'certain location', the field is 40/5 = 8 newtons per coulomb = <u>8 volts</u>
Answer:
Workdone = 465766038 Joules.
Explanation:
<u>Given the following data;</u>
Mass = 1167
Initial velocity = 10m/s
Final velocity =28m/s
To find the workdone;
We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.
Mathematically, it is given by the equation;
W = Kf - Ki
W = ½MVf² - ½MVi²
Substituting into the equation
W = ½(1167)*28² - ½(1167)*10²
W = ½ * 1361889* 784 - ½ * 1361889 * 100
W = 533860488 - 68094450
Workdone = 465766038 Joules.
Answer:
F=5449 N
Explanation:
Work done is a product of force and displacement ie
Work done, W, = Force*Displacement
Power, P, is Work done/Time
where P is power, W is work done, F is force, S is displacement and t is time
In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328 W
Since displacement/time is velocity, then
P=FV where V is velocity in m/s
Making F the subject
F=5449 N
