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amid [387]
2 years ago
13

Which of these are lost when the body repairs?

Physics
1 answer:
s2008m [1.1K]2 years ago
7 0

c. Sodium and potassium

<h3>What is called perspiration?</h3>

Perspiration, water given off by the intact skin, either as vapor by simple evaporation from the epidermis or as sweat, a form of cooling in which liquid actively secreted from sweat glands evaporates from the body surface.

When our body is sweating sodium and potassium is lost from the body along with water.

In order to maintain the integrity of the cells in the body ,Sodium and potassium are very important.

Sweat is also known as perspiration.

So,

By maintaining  the proper cell functioning and cell vitality optimum level of sodium and potassium should be maintained in the body.

Learn more perspiration about here:brainly.com/question/1237116

#SPJ1

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The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the
vovikov84 [41]

Answer:

F=3F_o(\frac{R_o}{R_{o2}})^2

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, Q_1 and Q_2 is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

F=\frac{kQ_1Q_2}{R^2}..................(1)

where k is the electrostatic constant.

We can make k the subject of formula  as follows;

k=\frac{FR^2}{Q_1Q_2}...........(2)

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from Q_o to 3Q_o and their distance of separation from R_o to R_{o2}, this also made the force between them to change from F_o to F_{o2}. Therefore as stated by equation (2), we can write the following;

\frac{F_oR_o^2}{Q_o*Q_o}=\frac{F_{o2}R_{o2}^2}{3Q_o*Q_o}.............(3)

Therefore;

\frac{F_oR_o^2}{Q_o^2}=\frac{F_{o2}R_{o2}^2}{3Q_o^2}.............(4)

From equation (4) we now make the new force F_{o2} the subject of formula as follows;

{F_oR_o^2}*{3Q_o^2}=F_{o2}R_{o2}^2*{Q_o^2}

Q_o then cancels out from both side of the equation, hence we obtain the following;

3{F_oR_o^2}=F_{o2}R_{o2}^2.............(4)

From equation (4) we can now write the following;

F_{o2}=\frac{3F_oR_o^2}{R_{o2}^2}

This could also be expressed as follows;

F_{o2}=3F_o(\frac{R_o}{R_{o2}})^2

3 0
3 years ago
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Answer:

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Explanation:

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Answer:

100Kg.m/s

Explanation:

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