Answer:
Tc = = 424.85 K
Explanation:
Data given:
D = 60 mm = 0.06 m

k = 50 w/m . k
c = 500 j/kg.k





HEAT FLOW Q is


= 47123.88 w per unit length of rod
volumetric heat rate





= 424.85 K
Answer:
3.25 ft/s
Explanation:
The crate is of =14-lb=m₁
The angle of inclination is = 40°=Ф
The initial velocity = 0.4 ft/s= v₁
Distance the crate will move is= 0.3 ft =d
The load pulling downwards is = 36 lb= m₂
Acceleration of the pulley, a= m₂g - m₁gsinФ / m₁+m₂ where g= 32.17 ft/s^2
a= 36*32.17 - 14*32.17*sin 40° / 14+36
a=17.37 ft/s^2
Apply the formula for final velocity
V₂²=V₁²+2ad
V₂²=0.4²+ 2*17.37*0.3
V₂²=10.582
V₂ =√10.582 = 3.25 ft/s
Answer:
The initial temperature will be "385.1°K" as well as final will be "128.3°K".
Explanation:
The given values are:
Helium's initial volume, v₁ = 6 m³
Mass, m = 1.5 kg
Final volume, v₂ = 2 m³
Pressure, P = 200 kPa
As we know,
Work, 
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
Now,
Gas ideal equation will be:
⇒ 
On putting the values. we get
⇒ 
⇒ 
⇒
(Initial temperature of helium)
and,
⇒ 
On putting the values, we get
⇒ 
⇒ 
⇒
(Final temperature of helium)