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Dafna11 [192]
2 years ago
10

Ill give brainliest! 1. Which item below describes a quick change to

Physics
2 answers:
Damm [24]2 years ago
8 0

Answer:

I believe it's <u>D. a glacier flattening the land below it​.</u>

stiks02 [169]2 years ago
7 0
The best answer to go with is b
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A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns
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Answer:

182.28 W

Explanation:

Here ,

m = 7.30 Kg

distance , d= 28.0 m

time , t = 11.0 s

average power supplied = change in potential energy/time

average power supplied = m×g×d/time

average power supplied = 7.30×9.81×28/11

average power supplied = 182.28 W

the average power supplied is  182.28 W

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2 years ago
Which of these describes a real image?
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Image from a far away object formed by a concave mirror

I have no idea but this is my best guess as a sophomore in college
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3 years ago
If the same experiment is repeated in different parts of the world by different scientists,
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The answer is D) The outcome of the experiment will be non observable
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3 years ago
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ASAP:
goldfiish [28.3K]

Explanation:

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5 0
2 years ago
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
3 years ago
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