Answer:
The specific heat capacity is q_{L}=126.12kJ/kg
The efficiency of the temperature is n_{TH}=0.67
Explanation:
The p-v diagram illustration is in the attachment
T_{H} means high temperature
T_{L} means low temperature
The energy equation :
= R* 
in(
/
)
The specific heat capacity:
=q_{h}*(T_{L}/T_{H})
q_{L}=378.36 * (400/1200)
q_{L}=378.36 * 0.333
q_{L}=126.12kJ/kg
The efficiency of the temperature will be:
=1 - (
/
)
n_{TH}=1-(400/1200)
n_{TH}=1-0.333
n_{TH}=0.67
so the 1st on is the one on the left, middle is right and the 3rd one is the right one
Answer:
The temperature of the core raises by 
every second.
Explanation:
Since the average specific heat of the reactor core is 0.3349 kJ/kgC
It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius
Thus reactor core whose mass is 
will require
energy to raise it's temperature by 1 degree Celsius in 1 second
Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by
Answer:
Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy, ΔPEg, is ΔPEg = mgh, with h being the increase in height and g the acceleration due to gravity.
Explanation:
You're Welcome.
Answer:
125.83672 seconds
Explanation:
P = Power of the horse = 1 hp = 746 W (as it is not given we have assumed the horse has the power of 1 hp)
m = Mass of professor = 103 kg
g = Acceleration due to gravity = 9.8 m/s²
h = Height of professor = 93 m
Work done would be equal to the potential energy
Power is given by
The time taken by the horse to pull the professor is 125.83672 seconds
