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3 years ago

What are some important factors to consider when choosing a warm-up before your workout?

Physics

2 answers:

TEA [102]3 years ago

6 0

Answer:

Prior to exercise, a proper warm-up of 10-15 minutes is extremely important to avoid injuries.

Don't go too hard in the beginning and boost your activity level slowly. A good indication of a proper warm-up is that you feel sweat on your body parts.
Don't overstretch right in the beginning as it can cause sore in your muscles and joints or stress fractures.
Take a break if you feel sick or fatigues and use other drinks along with water to replace electrolytes and body fluids.

Flauer [41]3 years ago

5 0

Answer:

Remember that the objective of warming up is avoiding problems when you are working out.

So the first thing that you need to have in mind is that you need to start slow, to prevent any possible harm.

You need to warm all your body, not only some parts of it.

when stretching, do not do it as you normally do, because it will relax your muscles, you should do dynamic stretching, this means that you keep your body in a position where there is no tension, and moving your body to generate tension, then relaxing again, then generating tension, and so on.

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Answer:

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

Explanation:

This is the complete question

A person will feel a shock when a current of greater than approximately 100μ A flows between his index finger and thumb. If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock compare in each scenario?

A) The voltage on dry skin needs to be 200 times smaller than the voltage on wet skin.

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

C) The voltage on dry skin is the same as the voltage on wet skin.

D) The voltage on dry skin needs to be 40,000 times larger than the voltage on wet skin.

Ohm's law states that electric current is proportional to voltage and inversely proportional to resistance.

the equation is written as

V = IR

Where V is the voltage

I is the current

R is the resistance

for this case, the current I is 100μ A = 100 x 10^16 A

resistance of wet skin = R

resistance of dry skin = 200R

for the wet skin, voltage will be

V = IR = $100*10^{-6} R$

for dry skin, voltage will be

V = IR = $100*10^{-6}*200R = 0.02R$

Comparing both voltages

$0.02R$ ÷ $100*10^{-6} R$ = 200

this means that the voltage on the wet skin should be 200 times lesser than the voltage on the dry skin or the voltage on the dry skin should be 200 times more than the voltage on the wet skin.

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