The correct answer is D. I alread took this test.
If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.
20 J/s = 10% of Supply
20 J/s = (0.1) x (Supply)
Divide each side by 0.1:
Supply = (20 J/s) / (0.1)
<em>Supply = 200 J/s </em>(200 watts)
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Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !
If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !
32.5 kg of air
Explanation:
To calculate the mass of the air, we use the density formula:
density = mass / volume
mass = density × volume
density of air = 1.3 kg/m³
volume = 5 × 3 × 2 = 25 m³
mass of the air = 1.3 kg/m³ × 25 m³
mass of the air = 32.5 kg
Learn more about:
density
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Answer:
reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year
Explanation:
given data
electricity consume = 14000 kWh
fuel consume = 900 gal
CO₂ produced of fuel = 26.4 lbm/gal
CO₂ produced of electricity = 1.54 lbm/kWh
oil and electricity usage = 21 percent
to find out
the reduction in the amount of CO₂ emissions
solution
we calculate the amount of CO₂ produce here that is
amount of CO₂ produce = ( electricity consume×CO₂ produce electricity + fuel consume × CO₂ consume fuel ) ........................1
put here value
amount of CO₂ produce = ( 14000 × 1.54 + 900 × 26.4 )
amount of CO₂ produce = 45320 lbm/yr
we know reduction is 21%
so
reduction in amount of CO₂ produced is
reduction in CO₂ produced = 45320 × 21%
reduction in CO₂ produced = 9517.2 lbm per year
so reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year
