HELP ME PLEASE If an object is placed under a force of 20 N, it accelerates at a rate of 5 m/s^2. If the force is increased to 5
2 answers:
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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
Answer:
Part a)
Part b)
Explanation:
As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same
So we will have
Part b)
By equation of kinetic energy we have