Answer:

Explanation:
In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.
Lets take




All external matting gears will rotates in opposite direction with respect to each other.
So the speed of gear third can be given as follows


Explanation:
The correct answers to the fill in the blanks would be;
1. Viscoelastic stress relaxation refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).
2. Viscoelastic creep refers to scenarios for which a polymer will permanently flow over time in response a constant applied stress.
The polymer whose properties have been mentioned above is commonly known as Kevlar.
It is mostly used in high-strength fabrics and its properties are because of several hydrogen bonds between polymer molecules.
The answer is choice C
Explanation:
As during construction ,the site is cleared for all debris before laying out the foundation. Even the sewer lines are dug out .
So it will be useful for the construction crews to connect the pipes to the sewer lines before the foundation is poured.
But usually the steps take in construction activity is:- first the site is cleared for the foundation to be poured and once the foundation wall is set , then all utilities , including plumbing and electrical activities are done.,
After this process is over, the city inspector comes to check whether the foundation has been laid down as per the code of construction.
Only after that the rest of the construction activity follows through.
The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.
<h3>What is power?</h3>
Power is the energy transferred per unit time.
Torque is find out by
P = 2πNT/60
10000 = 2π x 2000 x T / 60
T =47.74 N.m
The gear ratio Ne / Ns =4/1
Ns =2000/4 = 500
Ts =Ps x 60/(2π x 500)
Ts =190.96 N.m
Maximum shear stress τ = 16/π x (T / (d₀⁴ - d₁⁴))
τ max =T/J x D/2
where d₁ = 30mm = 0.03 m
d₀ = 30 +(2x 4) = 38mm =0.038 m
Substitute the values into the equation, we get
τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)
τ max = 28.98 MPa.
Thus, the maximum shear stress in the tube is 28.98 MPa.
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