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finlep [7]
3 years ago
13

A girl is running in a long distance race. As she runs, her respiration rate 10 points increases. Her body cells must process en

ergy at a faster rate to power her muscles as she runs. Is this action a response to internal stimuli or external stimuli?
A. internal stimuli
B. external stimili ​
Physics
1 answer:
Alik [6]3 years ago
4 0
A. Internal stimuli
Good luck
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In phase would mean both waves are at a positive peak, out of phase would mean one is at a positive whilst the other is at a negative. Out of phase would mean the waves cancel each other out
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Is the classification for an instrument that produces sound whne a string or strings stretched between two points is plucked?
Yuki888 [10]
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5 0
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NASA developed a reflective foil barrier to help shield spacecraft from heat transfer in space. These reflective barriers are no
morpeh [17]

Answer: Radiation

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3 0
3 years ago
A +0.2 µC charge is in an electric field. What happens if that charge is replaced by a +0.4 µC charge?
AleksandrR [38]

Explanation:

It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.

So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.

                 U = qV

Since, the electric potential is directly proportional to the charge. Hence, when 0.2 \mu C tends to replaced by 0.4 \mu C then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.

Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.

4 0
3 years ago
Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
3 years ago
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