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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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A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur
Dmitriy789 [7]

Answer:

d = 2.54 [m]

Explanation:

Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

E_{k1}+W=E_{k2}\\

where:

W = work [J]

Ek1 = kinetic energy at initial state [J]

Ek2 = kinetic energy at the final state = 0.

We must remember that kinetic energy can be calculated by means of the following expression.

\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]

We know that work is defined as the product of force by distance.

W=F*d

where:

F = force [N]

d = distance [m]

But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.

f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]

Now solving the equation for the work.

d=W/F\\d = 50/19.62\\d = 2.54[m]

4 0
3 years ago
A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 so
hodyreva [135]

Answer:A) Risk(R)= $1000

B) There is justification for spending an additional cost of $100 to prevent a corrosion whose consequence in monetary terms is $1000

Explanation:R= Risk,

P=Probability of failure

C= Consequence of failure

Mathematically, R=P ×C

10 out of 1000 carbon-steal products failed

Probability of failure= 10/1000 =0.01

The consequence of failure by corrosion given in monetary term =$100,000

Risk of failure = 0.01 × $100,000

R=$1000

4 0
3 years ago
Two balls of mass 0.09 kg hang on strings attached to the same point on the ceiling. The balls are given charges Q that cause th
telo118 [61]

Answer:

Q = 6.33μC

Explanation:

To find the value of the charge Q you take into account both gravitational force and electric force over each ball. By symmetry you can use the fact that both balls experiences the same forces. Hence you only take into account the forces for one ball for the x component and y component:

-Mg+Tcos\theta=0\\\\F_e-Tsin\theta=0

M: mass of the ball = 0.09kg

T: tension of the string

F_e: electric force between charges

angle = 45°

The electric force is given by:

F_e=k\frac{Q^2}{r^2}

Q: charge of the balls

r: distance between balls = 2m

You divide both equation in order to eliminate the tension T:

tan\theta=\frac{F_e}{Mg}=k\frac{Q^2}{Mgr^2}

By doing Q the subject of the formula and replacing you obtain:

Q=\sqrt{\frac{tan\theta Mgr^2}{k}}=\sqrt{\frac{tan45\°(0.09kg)(2m)^2}{(8.89*10^{9}Nm^2/C^2)}}=6.33*10^{-6}C=6.33\mu C

hence, the charge of the balls is 6.33μC

4 0
3 years ago
Is a balance scale used to describe how much matter is in an object
gtnhenbr [62]
It's used to measure the mass or weight of an object
7 0
3 years ago
Read 2 more answers
which of the following would be best to do if you were inside a car and a power line fell on the car​
NARA [144]

Answer:

Do not move stay in your car and wait for someone from the power company to come and help

Explanation:

Plz vote my answer as the brainiest, i rlly need it! hope this helps!

6 0
3 years ago
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