Work = Force x Distance = 500 x 4 = 2000 Nm = 2000 J
Angstrom = 10^-10 m
for nucleus size are used fermi (femtometer 10^-15 m )
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:
a) 1450watts
b) 564watts
c) 1.11
Explanation:
Power consumed = IV
I is the current rating
V is the operating voltage
If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;
a) For blow-dryer
Operating voltage = 120V
Its current rating = 12A
Power consumed = IV
= 120×12
= 1440watts
b) For vacuum cleaner:
Operating voltage is the same as that of blow dryer = 120V
Its current rating = 4.7A
Power consumed = IV
= 120×4.7
= 564watts
c) Energy used = Power consumed × time taken
Energy used = Power × time
Energy used by blow dryer = 1440×20×60
= 1,728,000Joules
Energy used up by vacuum cleaner = 564×46×60
= 564×2760
= 1,556,640Joules
Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11
