Answer: The exit temperature of the gas in deg C is .
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa,
We know that for an ideal gas the mass flow rate will be calculated as follows.
or, m =
=
= 10 kg/s
Now, according to the steady flow energy equation:
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
=
Therefore, we can conclude that the exit temperature of the gas in deg C is .
Answer:
D=41.48 ft
Explanation:
Given that
y=0.5 x²
Vx= 2 t
We know that
At t= 0 ,x=0
At t= 3 s
x= 9 ft
When x= 9 ft then
y= 0.5 x 9² ft
y= 40.5 ft
So distance from origin is
x= 9 ft ,y= 40.5 ft
D=41.48 ft
Vx= 2 t
At t= 3 s , x= 9 ft
y=0.5 x²
y=0.5 x²
Given that
Answer:
Computer program
Explanation:
I use Revit and its way better to do all that you can see 2D 3D the measurements and its super easy to use hope this helps
Answer:
transmission bandwidth required is very large.
Explanation:
Answer:
591.3
Explanation:
99.19 + (1.85 × 266) = 591.29
rounded = 591.3