Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.
Answer:
The magintude of the acceleration for both objects is 
Explanation:
Drawing a free body diagram on the two boxes we can analyze the system more easily.
we can take the acceleration going up as positive for reference purposes.
for mA let's suppose that is ascending so:

and for mB (descending):


because the two boxes has the same acceleration because they are attached together:

So the magintude of the acceleration for both objects is 
Answer:
When you have to do an English-Metric (SI) length conversion, and you already know the English units of length (miles, yards, feet, inches, etc.), all you need to remember is one simple relationship, and you can readily convert any length in the SI system, to the equivalent length in the other.
1 foot (ft) = 0.3048 meters (m)
BIn this case you need your answer in inches. You (hopefully) know there are 12 inches in a foot, so you just do the following:
1 inch (in) = 1/12 ft = 0.3048/12 m = 0.0254 m
Answer:
Ae/A* = 1.115
Explanation:
Let the reservoir pressure be 
Let the exit pressure be 
Ratio of reservoir pressure and exit pressure

= 3.182
For the above value of pressure ratio
Obtain the area ratio from the isentropic flow table
Ae/A* = 1.115
The value of pressure ratio is Ae/A* = 1.115