To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as

Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as

where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then

Resolving for F,

PART B) We need here to apply the shear strength equation, then



In such a way that the material is more resistant to tensile strength than shear force.
Answer:
"A turbine takes the kinetic energy of a moving fluid, air in this case, and converts it to a rotary motion. As wind moves past the blades of a wind turbine, it moves or rotates the blades. These blades turn a generator."
Answer:
is the time taken by the car to accelerate the desired range of the speed from zero at full power.
Explanation:
Given:
Range of speed during which constant power is supplied to the wheels by the car is
.
- Initial velocity of the car,

- final velocity of the car during the test,

- Time taken to accelerate form zero to 32 mph at full power,

- initial velocity of the car,

- final desired velocity of the car,

Now the acceleration of the car:



Now using the equation of motion:


is the time taken by the car to accelerate the desired range of the speed from zero at full power.
<u>Given data</u>
Source temperature (T₁) = 177°C = 177+273 = 450 K
Sink temperature (T₂) = 27°C = 27+273 = 300 K
Energy input (Q₁) = 3600 J ,
Work done = ?
We know that, efficiency (η) = Net work done ÷ Heat supplied
η = W ÷ Q₁
W = η × Q₁
First determine the efficiency ( η ) = ?
Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)
= 33.3% = 0.333
Now, Work done is W = η × Q₁
= 0.33 × 3600
<em> W = 1188 J</em>
<em>Work done by the engine is 1188 J</em>