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exis [7]
3 years ago
9

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics
3 answers:
xxMikexx [17]3 years ago
8 0
14-6 =8
8/4= 2m/s per second
Leto [7]3 years ago
6 0

accn=change in vel/time

8/4=2m/s/s

jeka943 years ago
4 0

accn=change in vel/time

8/4=2m/s/s

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You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength
GREYUIT [131]

Answer:

Period of the signal.

Explanation:

So, this question is all about a concept in physics or astronomy which is called or known as Radiation Astronomy and Galactic Nuclei that are active. This concept talks most about Quasars; a powerful radiating object which derives its power from black holes.

When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.

Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.

NB: not the amplitude but the period.

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3 years ago
Your chances of getting into a collision when talking on a cell phone _________: A. Double B. Triple C. Quadruple D. Remain the
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C. Quadruple

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¨Drivers who are talking on the phone, even on a hands-free device, are up to four times more likely to be involved in a crash.¨

I hope this helps! Have a great day!

3 0
2 years ago
Which kind of mirror can produce real images?<br><br><br> A)concave<br><br> B)convex<br><br> C)flat
Anton [14]

Answer:

C

Explanation:

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3 years ago
What question can a student BEST answer when comparing and contrasting the models?
OverLord2011 [107]

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4 0
2 years ago
A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperatur
Anna35 [415]

Answer:

\dot{W_{H} } = 4244.48 Btu/h

Explanation:

Temperature of the house, T_{H} = 70^{0} F

Convert to rankine, T_{H} = 70^{0}+ 460 = 530 R

Heat is extracted at 40°F i.e T_{L} = 40^{0}F  = 40 + 460 = 500 R

Calculate the coefficient of performance of the heat pump, COP

COP = \frac{T_{H} }{T_{H} - T_{L}  } \\COP = \frac{530 }{530 - 500  }\\ COP = \frac{530}{30} \\COP = 17.67

The minimum power required to run the heat pump is given by the formula:

\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\...............(*)

Where the heat losses from the house, \dot{Q_{H} } = 75,000 Btu/h

Substituting these values into * above

\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h

3 0
2 years ago
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