Answer:
1. 
2. 
Explanation:
1.
Given:
- height of the window pane,

- width of the window pane,

- thickness of the pane,

- thermal conductivity of the glass pane,

- temperature of the inner surface,

- temperature of the outer surface,

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

here:
A = area through which the heat transfer occurs = 
dT = temperature difference across the thickness of the surface = 
dx = t = thickness normal to the surface = 


2.
- air spacing between two glass panes,

- area of each glass pane,

- thermal conductivity of air,

- temperature difference between the surfaces,

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>



Answer: a) W(earth) = 935.62 lbs
b) Mass of rocks in slugs = 29.06 slugs
Explanation:
a) From Newton's law, W = mg. Whether on the moon or on earth. Although, the mass of the rocks everywhere is the same, that is, mass of rocks on the moon = mass of rocks on earth.
W(moon) = mg(moon)
W(moon) = 154 lbs
g(moon) = 5.30 ft/s2
m = W(moon)/g(moon) = 154/5.3 = 29.06 lb.s2/ft
W(earth) = m g(earth)
g(earth) = 32.2 ft/s2
W(earth) = 29.06 × 32.2 = 935.62 lbs.
b) A slug = 1 lb.s2/ft, therefore the mass of the rocks in slugs is 29.06 slugs.
QED!
Answer:
a) the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Explanation:
Given the data in the question;
Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in
Diameter d = 3/16 in
Unit weight w = 490 pcf
First we determine the area of the wire;
A = π/4 × d²
we substitute
A = π/4 × (3/16)²
A = 0.0276 in²
Next we get the Volume
V = Area × Length of wire
we substitute
V = 0.0276 × 4800
V = 132.48 in³
Weight of the steel wire will be;
W = Unit weight × Volume
we substitute
W = 490 × ( 132.48 / 12³ )
W = 490 × 0.076666
W = 37.57 lb
a) the maximum tensile stress due to the weight of the wire;
σ
= W / A
we substitute
σ
= 37.57 / 0.0276
= 1361.23 psi
Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi
Maximum load P that the wire can safely support its lower end will be;
P = ( σ
- σ
)A
we substitute
P = ( 24000 - 1361.23 )0.0276
P = 22638.77 × 0.0276
P = 624.83 lb
Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Answer:
The p-value of the hypothsis 0.00135
Explanation: