I don’t know sorry ;khbadkhb didhwbck( khwdicdwbihwd
On sources it says it would just be the super giant star
Answer:
The temperature of cold reservoir should be 246.818 K for efficiency of 35%
Explanation:
In first case we have given efficiency of Carnot engine = 26 % = 0.26
Temperature of cold reservoir 
We know that efficiency of Carnot engine is given by



For second Carnot engine efficiency is given as 35% = 0.35
And temperature of hot reservoir is same so 
So 

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%
Answer:
(a) 
(b) 5220 j
(c) 1740 watt
(d) 3446.66 watt
Explanation:
We have given mass m = 290 kg
Initial velocity u = 0 m/sec
Final velocity v = 6 m/sec
Time t = 3 sec
From first equation of motion
v = u+at
So 
(a) We know that force is given by
F = ma
So force will be 
(b) From second equation of motion we know that

We know that work done is given by
W = F s = 580×9 =5220 j
(c) Time is given as t = 3 sec
We know that power is given as

(d) Time t = 1.5 sec
So 