Answer:
(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%
Explanation:
Using the formula that relate heat and work from the thermodynamic theory as:
solving to Q_out we get:
this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us:
where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:
Answer:
The minimum speed must the car must be 13.13 m/s.
Explanation:
The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.
We know that, mg be the weight of car and rider, which is equal to the centripetal force.

So, the minimum speed must the car must be 13.13 m/s.
Answer:
660 centimeters
Explanation:
There are 100 cm in 1 m. To convert from m to cm, multiply by 100.

There are 660 cm in 1 m.
Answer:-600N
Explanation:
Mass(m)=1000kg
Initial velocity(u)=25m/s
Final velocity(v)=12.5m/s
Time(t)=22 seconds
v=u+axt
12.5=25+ax22
collect like terms
12.5-25=22a
-12.5=22a
Divide both sides by 22
-12.5/22=22a/22
-0.6=a
Acceleration=-0.6
Force=mass x acceleration
Force=1000 x -0.6
Force=-600