Answer:1200
Explanation:
Given data
Upper Temprature
Lower Temprature 
Engine power ouput
Efficiency of carnot cycle is given by





rounding off to two significant figures

I believe that the price will rise.
Hope this helps!
Answer:
Explanation:
Force = mass * acceleration.
2 seconds,,,,,,,,,,,,,,,,,,,,,,,
Answer:
The change in current at
is 
Explanation:
From the question we are told that
The resistance is 
The current is 
The change in voltage with respect to time is 
The change in resistance with time is 
According to ohm's law

differentiating with respect to time using chain rule

substituting value at R = 456

