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Fiesta28 [93]
3 years ago
8

John stretched while his muscles were cold. Which negative effect could occur?

Physics
2 answers:
Ede4ka [16]3 years ago
5 0

Answer:

The answer for Plato is C.

Genrish500 [490]3 years ago
3 0
<h3><u>Answer;</u></h3>

C) He could injure or pull a muscle.

<h3><u>Explanation</u>;</h3>
  • <em><u>When muscles are stretched the muscle fibers are temporary lengthened. Muscles have a unique ability to undergo contraction and lengthening since they are elastic.</u></em>
  • <em><u>When the muscles are warm they are more elastic, therefore, warming up the muscles is very important for the purpose of avoiding injuries.</u></em> Ir is important to engage in moderate cardiovascular warm-up prior to stretching, which increases the blood flow to the active and thus avoiding injuries to the muscles.
  • Stretching muscles when they are cold may bring injuries or cause muscle pulls, which are painful.
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The same amount of substance was added to four beakers of water. The treatments were placed in the chart.
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Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper
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Explanation:

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             R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}  

where,   L_{I} = length of the iron bar

             k_{I} = thermal conductivity of iron

             A_{I} = Area of cross-section for the iron bar

Thermal resistance for copper (R_{c}) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where,  L_{c} = length of copper bar

             k_{c} = thermal conductivity of copper

            A_{c} = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

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       Q = \frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

  = \frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}

           = 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

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Here, T is 1 second so, power conducted is equal to heat transferred.

So,           P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

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