Given 1 inch ≡ 2.54 cm and 1 foot ≡
1 answer:
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Answer:
The work done on the Frisbee is 1.36 J.
Explanation:
Given that,
Mass of Frisbee, m = 115 g = 0.115 kg
Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground
Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,
So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.
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: