The pressure inside the can upon cooling is 0.4 atm.
<u>Explanation:</u>
Given -
Initial Temperature, T1 = 908°C = 908 + 273 K = 1181 K
Final Temperature, T2 = 208°C = 208 + 273 K = 481 K
Pressure upon cooling, P2 = ?
Using Gay Lussac's law:
P1/T1 = P2/T2
P2 = P1 X T2 / T1
P2 = 1 atm X 481 / 1181
P2 = 0.4 atm
Therefore, the pressure inside the can upon cooling is 0.4 atm.
I believe number 4 I could be wrong but I think it’s 4
Solution :
Given :
The power of the air‐conditioning (AC) unit is , W = 0.434 kW
The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by = 6.22
Therefore he heat removed is given by , 

Now if the electricity is valued at 0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24
= 6.48
Therefore the operating cost = $ 6.48 for 24 hours.
Answer:
W = 11,416.6879 N
L ≈ 64.417 cm
Explanation:
The maximum shear stress,
, is given by the following formula;

= 1 cm = 0.01
h = 29 cm = 0.29 m
= 25 cm = 0.25 m
b = 15 cm = 0.15 m
= The centroidal moment of inertia
= 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴
Substituting the known values gives;

= 1.2257083
× 10⁻⁴ m⁴
From which we have;

Which gives;
W = 11,416.6879 N

= 1500 N/cm² = 15,000,000 N/m²
= 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²
From Which we have;


L ≈ 0.64417 m ≈ 64.417 cm.
Answer:
Step 1
Given
Diameter of circular grill, D = 0.3m
Distance between the coal bricks and the steaks, L = 0.2m
Temperatures of the hot coal bricks, T₁ = 950k
Temperatures of the steaks, T₂ = 5°c
Explanation:
See attached images for steps 2, 3, 4 and 5