The modulus of elasticity is 28.6 X 10³ ksi
<u>Explanation:</u>
Given -
Length, l = 5in
Force, P = 8000lb
Area, A = 0.7in²
δ = 0.002in
Modulus of elasticity, E = ?
We know,
Modulus of elasticity, E = σ / ε
Where,
σ is normal stress
ε is normal strain
Normal stress can be calculated as:
σ = P/A
Where,
P is the force applied
A is the area of cross-section
By plugging in the values, we get
σ = 
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε = 
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:

Thus, modulus of elasticity is 28.6 X 10³ ksi
Answer:
Jordan has more green paints
Explanation:
Given


Required
Which paint does he have more?
For better understanding, it's better to convert both measurements to decimal.
For the green paint:


For the blue paint:


By comparison:

<em>This means that Jordan has more green paints</em>
Answer:
DIAMETER = 9.797 m
POWER = 
Explanation:
Given data:
circular windmill diamter D1 = 8m
v1 = 12 m/s
wind speed = 8 m/s
we know that specific volume is given as

where v is specific volume of air
considering air pressure is 100 kPa and temperature 20 degree celcius

v = 0.8409 m^3/ kg
from continuity equation





mass flow rate is given as


the power produced ![\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]](https://tex.z-dn.net/?f=%5Cdot%20W%20%3D%20%5Cdot%20m%20%5Cfrac%7B%20V_1%5E2%20-%20V_2%5E2%7D%7B2%7D%20%3D%20717.3009%20%5B%5Cfrac%7B12%5E2%20-%208%5E2%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2Fs%5E2%7D%5D)

Snap rings, and bearings can be used to keep a gear on a shaft, hope this helps!!