Answer:
THE ANSWERS IS A MODERN ANSWERS IS THE BIOLOGY WAY
Explanation:
Answer:
he distance to a star can be found by comparing its luminosity to its apparent brightness.
Explanation: A star that is far away will be very dim compared to its luminosity. So it proves that the ratio of its luminosity to apparent brightness will be greater than for a nearby star.
Answer:
Sun > Producer > Primary Consumer > Secondary Consumer > Tertiary Consumer > (sometimes Quaternary Consumer >) Decomposer
Explanation:
Answer:
- The annual cost for electricity to power a lampost for 7.00 h per day with a 100 W incandescent bulb is $ 40.89.
- The annual cost for electricity to power the same lampost for the same time with an energy efficient 25 W fluorescent bulb is $ 10.22.
So, you can save $ 40.89 - $ 10.22 = $ 30.67 per each lampost if you replace the incadescent bulb with the energy efficient fluorescent bulb.
Explanation:
kW.h means kilowatt.hour and is a unit of energy derived from the definition of power.
The definition of power is energy per time:
Thus, you get:
For the units, you get:
- [Energy] = [Power] [time] = KW.h.
<u>1. Calculate the energy used by the 100 W incandescent light bulb that is used 7.00 h per day, in a year (365 days).</u>
- E = Power × time = 100 W × 7.00 h / day × 365 day / year × 24 h= 255,500 W.h/year
- Divide by 1,000 to convert to kW: 255.5 kW.h/year
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<u>2. Calculate the energy used by the 25 W flourescent bulb:</u>
- E = 25 W × 7.00 h / day × 365 day / year × 24 h= 63,875 W.h/year
- Divide by 1,000 to convert to kW: 63.875 kW.h/year
<u>3. Now, you can calculate both costs:</u>
- Cost per year = cost per kW.h times number of kW.h per year
<u>100 W bulb:</u>
<u />
- $ 0.16/ kW.h × 255.55 kW.h/ year = $ 40.89 / year
<u>25 W bulb:</u>
- $ 0.16/ kW.h × 63.875 kW.h/ year = $ 10.22 / year