Increases exponentially is your correct answer
Answer:
<em><u>option</u></em><em><u> (</u></em><em><u>C)</u></em><em><u> </u></em><em><u>is </u></em><em><u>right</u></em><em><u> answer</u></em>
Explanation:
I think it's helps you
First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.
Answer:
Temperature at the exit = 
Explanation:
For the steady energy flow through a control volume, the power output is given as

Inlet area of the turbine = 
To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.
Assuming Argon behaves as an Ideal gas, we have the specific volume 
as


for Ideal gasses, the enthalpy change can be calculated using the formula

hence we have


<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>
evaluating the above equation, we have 
Hence, the temperature at the exit = 