Answer:

is time required to heat to boiling point form initial temperature.
Explanation:
Given:
initial temperature of water, 
time taken to vapourize half a liter of water, 
desity of water, 
So, the givne mass of water, 
enthalpy of vaporization of water, 
specific heat of water, 
Amount of heat required to raise the temperature of given water mass to 100°C:



Now the amount of heat required to vaporize 0.5 kg of water:

where:
mass of water vaporized due to boiling


Now the power rating of the boiler:



Now the time required to heat to boiling point form initial temperature:


Answer 41.7 years old
I answer this question already and it was correct!
Moving fan has rotational kinetic energy
Non moving fan has no energy since it is in rest
Answer:
1200 meters
Explanation:
there are 60 seconds in a minute times 2 is 120 ten times 120 is 1200
The power that the light is able to utilize out of the supply is only 0.089 of the given.
Power utilized = (0.089)(22 W)
= 1.958 W
= 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J.