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Furkat [3]
3 years ago
7

How does exercise help you body devolpe

Physics
1 answer:
seropon [69]3 years ago
8 0
An exercise strains the body and in response, your body heals/repairs itself and makes it better


example:
pushups strain the mucles in your arms and chest.
after the exercise, they might be sore. that's because they are damaged and need to heal. when they heal, they will be stronger and more durable
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Please help asap I need this rn
den301095 [7]
It’s the third one .
6 0
3 years ago
If a fish is trying to capture an insect hovering above the surface of the water – how will it jump to catch it? Will it aim abo
Viktor [21]

Answer:

It will have aim at a point "below" the insect.

From the insect's point of view, the fish will appear to be shallower than it actually is because a ray of light from the insect to the fish will be bent "towards" the normal when the ray enters the water

5 0
2 years ago
A body is placed on a rough inclined plane. Why does the frictional force decrease with the increase of angle of inclination
kondor19780726 [428]
The frictional force is directly proportional to the force that is perpendicular on the surface.

When the body is placed on a horizontal level with zero inclination, the only force acting on the body is the gravitational force which always pulls the body down. The gravitational force, in this case, is the perpendicular force to the surface. Accordingly, this entire force is used to generate friction

Now as the inclination of the surface increases, the gravitational force is no longer the perpendicular force of the body, its value decreases, which means only a part is used to generate frictional force. Consequently, frictional force decreases.

When the inclination reaches 90 degrees, the gravitational force does not act along the normal and accordingly, no friction force is generated.

7 0
3 years ago
I dolt understand what to do next please help
lidiya [134]
Divide 24 by 12.
24/t = 12
24/12 = t
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6 0
3 years ago
Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at
dybincka [34]

Answer:

The intensity at a spot 71 m away is 4.02*10^{-5}  Wm^{-2}

Explanation:

Given:

Initial intensity,I_{1}=3.0*10^{-4} Wm^{-2} at a distance, d_{1} = 26 m

Required:

New intensity, I_{2} =? at a distance, d_{2} = 71 m

Using the inverse square law,

I ∝ \frac{1}{d^{2} }

⇒I_{1}I_{1}d_{1}^{2}  =I_{2}d_{2}^{2}

I_{2} =\frac{I_{1} d_{1}^{2}  }{d_{2}^{2}} =\frac{3.0*10^{-4}*26^{2}  }{71^{2} } \\×

I_{2}=4.02*10^{-5}  Wm^{-2}

Thus, the intensity at a spot that is 71 m away is 4.02*10^{-5}  Wm^{-2}

5 0
3 years ago
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