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RUDIKE [14]
3 years ago
11

Part C Let’s start the analysis by looking at your “extreme usage” cases. Compare the two cases in detail—low usage period versu

s high usage period. Discuss differences between the two as well as any surprises. Things you should cover in your discussion: How much difference was there in average power usage (avg. kW) between the low-usage and high-usage time periods? What might have been running during the low-usage period that used energy? Identify likely “stealth” energy users that you could not turn off during the low-usage period. What do you suppose contributed most to the usage during the high-usage period?
Physics
1 answer:
Serga [27]3 years ago
6 0

Answer:

Day 7 DataUsage notes (since last reading)day & datetimekWh readingkWh usedhours elapsedavg. kW usedb.Usage Extremes: Data CollectionFor this experiment, you’ll measure electrical usage during a time period when you expect to havevery light electrical usage (for instance, while you’re asleep at night or during the day when no oneis at home). Likewise you’ll measure electrical usage during a time period when you expect to have heavier than average electrical usage. This time period might be in the evening, when lights and other appliances are on. Both of these time periods should be at least 4 hours long, to increase the accuracy of your results. Record your results in the tables below for each situation. For each time period, you’ll need to takean initial and a final reading.Type your response here:Low Usage - Initial Readingday & datetimekWh readingLow Usage - Final ReadingEnergy Usage Notesday & datetimekWh readingkWh usedhours elapsedavg. kW usedHigh Usage - Initial Readingday & datetimekWh reading4

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Alina [70]

Answer:

V = 25.3 , θf = 36.7° below the horizontal

Explanation:

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The earth and the moon exert forces on each other which forces is greater? explain
Helen [10]

Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.

An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).

The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:

ag=G(MEarth+MMoon)/r2

Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:

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Where T is the period. Since the two accelerations have to be equal, we obtain:

(4 pi2 r) /T2=G(MEarth+MMoon)/r2

Which implies:

r3/T2=G(MEarth+MMoon)/4 pi2=const.

This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.

This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.


8 0
3 years ago
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3 years ago
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