Answer:
In terms of energy conversion/transformation the generator that produces 40 kilowatts of power will consume twice as much fuel than a generator that produces 20 watts because the more electrical power produced by an electrical generator, the more fuel is consumed. The amount of electrical energy produced is directly proportional to the fuel consumed.
Explanation:
Energy transformation is the conversion of energy from one form to another. Their are various forms of energy namely kinetic energy, electric energy, mechanical energy , chemical energy , solar energy, heat energy , nuclear energy. These energy can be transformed from one form to another . Example our body convert chemical energy through food we eat to mechanical energy.
The generator transforms mechanical energy into electrical energy . The kilowatts tells you the amount of power the generator produces .The energy is measured in kilowatts per hour which is the power used over a period of time.
In terms of energy conversion/transformation the generator that produces 40 kilowatts of power will consume twice as much fuel than a generator that produces 20 watts because the more electrical power produced by an electrical generator, the more fuel is consumed. The amount of electrical energy produced is directly proportional to the fuel consumed. So far the generator conversion efficiency are equal for both generators the generator with the higher power generation will consume more fuel per hour.
Answer:
What is percentage modulation in AM?
The percent modulation is defined as the ratio of the actual frequency deviation produced by the modulating signal to the maximum allowable frequency deviation.
Answer:
A) approximate alkalinity = 123.361 mg/l
B) exact alkalinity = 124.708 mg/l
Explanation:
Given data :
A) determine approximate alkalinity first
Bicarbonate ion = 120 mg/l
carbonate ion = 15 mg/l
Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61
= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3
B) calculate the exact alkalinity of the water if the pH = 9.43
pH + pOH = 14
9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57
[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l
[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l
[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l
therefore the exact alkalinity can be calculated as
= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )
= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )
= 124.708 mg/l
