Architects must have a professional bachelor's or master's degree in architecture from a program that has been accredited by the National Architectural Accrediting Board, and a state license.
Answer:
2.65 MPa
Explanation:
To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

<u>Where:</u>
p: is the gage pressure = 108 kPa
r: is the inner radius of the ball
t: is the thickness = 3 mm
Hence, we need to find r, as follows:

<u>Where:</u>
d: is the outer diameter = 300 mm

Now, we can find the normal stress (σ) in the wall of the basketball:
Therefore, the normal stress is 2.65 MPa.
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Answer:
t=5.5 mm
Heat dissipation per unit length = 90.477 W/m
Explanation:
Given that
Diameter d = 5 mm ⇒r = 2.5 mm
Conductivity of insulated material K = 0.16 W/mK
Heat transfer coefficient = 20 
When thickness reaches up to critical radius of insulation then heat dissipation will be maximum
We know that critical radius of insulation of wire is given as follow

Now by putting the values


So the thickness of insulation
t=8-2.5 mm
t=5.5 mm
As we know that heat transfer due to convection given as follows
Q = hAΔ T
Q=20 x 2 x π x 0.008 x (120-30)
Q = 90.477 W/m
So heat dissipation per unit length = 90.477 W/m
Answer:
A) 209.12 GPa
B) 105.41 GPa
Explanation:
We are given;
Modulus of elasticity of the metal; E_m = 67 GPa
Modulus of elasticity of the oxide; E_f = 390 GPa
Composition of oxide particles; V_f = 44% = 0.44
A) Formula for upper bound modulus of elasticity is given as;
E = E_m(1 - V_f) + (E_f × V_f)
Plugging in the relevant values gives;
E = (67(1 - 0.44)) + (390 × 0.44)
E = 209.12 GPa
B) Formula for upper bound modulus of elasticity is given as;
E = 1/[(V_f/E_f) + (1 - V_f)/E_m]
Plugging in the relevant values;
E = 1/((0.44/390) + ((1 - 0.44)/67))
E = 105.41 GPa