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Oksana_A [137]
3 years ago
7

Kristina works out seven days a week. Lately, she has been tired, and her body aches. If she is overtraining, which training pri

nciple should Kristina consider before continuing her program?
a. overload
b. recovery
c. transfer
d. variation
Physics
1 answer:
sertanlavr [38]3 years ago
8 0

Answer: Option (b) is the correct answer.

Explanation:

As Kristina is over training, therefore, there is pain in her muscles because human body also requires rest in order to work or function properly.

That is why, gym trainers suggest to give one day off in a week for your workout sessions so that body muscles should recover.

Thus, we can conclude that if Kristina is over training, then recovery training principle should Kristina consider before continuing her program.

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What is the ratio of the sun’s gravitational pull on Mercury to the sun’s gravitational pull on the earth?
Marta_Voda [28]

Answer:

The answer is \frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709. Let's learn why.

Explanation:

Newton's law of universal gravitation says;

F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}

Here G is a universal gravitational <u>constant</u> and is measured experimentally.

Sun's gravitational pull on mercury is:

F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }

Therefore F_{Sun-Mercury} = Gm_{sun} 98,4366

Sun's gravitational pull on Earth is:

F_{Sun-Earth} =G.\frac{m_{sun} 5,97.10^{24} }{(1,50.10^{11}) ^{2}}

Therefore F_{Sun-Earth} =Gm_{sun} 265,33

As a result;

\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709

4 0
4 years ago
The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which
spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  \frac{1}{v} -\frac{1}{u} =\frac{1}{f}

On putting estimate values, we get

⇒  \frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}

⇒  \frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}

⇒  v=1.94 \ cm

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

m=\frac{v}{u} and m=\frac{h_{1}}{h_{0}}

⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

⇒  {h_{1}}=1.16 \ mm

8 0
3 years ago
The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21ms. The total mass of the train is 825 g.
Rudiy27

Explanation:

Average power = change in energy / change in time

P = ΔE / Δt

P = (½ mv²) / t

P = (½ (0.825 kg) (0.620 m/s)²) / (0.021 s)

P = 7.55 Watts

7 0
3 years ago
The moment of a force is calculated from the product of the ———— and the———— distance from the line of action of the force to th
bixtya [17]

Answer: It is the product of the (force)multiplied by the (perpendicular) distance from the line of action of the force to the pivot

Explanation:

7 0
3 years ago
How does the input distance of a single fixed pulley compare to the out- put distance?
ololo11 [35]

A pulley is another sort of basic machine in the lever family. We may have utilized a pulley to lift things, for example, a banner on a flagpole.

<u>Explanation:</u>

The point in a fixed pulley resembles the support of a lever. The remainder of the pulley behaves like the fixed arm of a first-class lever, since it rotates around a point. The distance from the fulcrum is the equivalent on the two sides of a fixed pulley. A fixed pulley has a mechanical advantage of one. Hence, a fixed pulley doesn't increase the force.

It essentially alters the direction of the force. A moveable pulley or a mix of pulleys can deliver a mechanical advantage of more than one. Moveable pulleys are appended to the item being moved. Fixed and moveable pulleys can be consolidated into a solitary unit to create a greater mechanical advantage.

4 0
3 years ago
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