Please help, last question i can ask, brainest if correct

Physics

1 answer:

Ugo [173]3 years ago

7 0

Answer:

C. Athletic performance peaks during teen years

Explanation:

According to me this is the right answer because in A it is actually men’s athletic ability that peeks later than women. In B, it is kinda obvious that is not true.

Hope this helps!

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A wave can be best defined as P.S pls help

Sidana [21]

Answer:

It should be A. Disturbance that travels through a medium or space, transmitting energy from one point to another.

I hope this helped you :)

4 0

3 years ago

Compare and contrast sound waves and electromagnetic waves. Any bs answers will be reported. Please hurry!!!

Trava [24]

Answer:

They are both forms of energy. One has to do with hearing and the other a little bit of light.

Explanation:

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3 years ago

A 23.8 kg space rock is pushed with a force and accelerates at 8.97 m/s/s. How much force was applied?

kotegsom [21]

m=23.8kg a=8.97m/s^2 Fnet=? Fnet=ma=(23.8kg)(8.97m/s^2)=213.486N

5 0

3 years ago

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be $v_{AT}=v_A-v_T$ (the train must see the car advancing at a lower speed), where $v_A$ is the speed of the automobile and $v_T$ the speed of the train.

So we have $v_{AT}=(95km/h)-(75Km/h)=20Km/h$ .

So the train (anyone in fact) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s$

And in that time the car would have traveled (relative to the ground):

$d=v_At=(95Km/h)(0.065h)=6.175Km$

If they are traveling in opposite directions, we have to do all the same but using $v_{AT}=v_A+v_T$ (the train must see the car advancing at a faster speed), so repeating the process: