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3 years ago

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3. How does temperature affect a football players performance?

1 answer:

castortr0y [4]3 years ago

7 0

Hello!

3. How does temperature affect a football players performance?

Independent Variable: Temperature

<em>Temperature is the independent variable because that is what you are testing.</em>

Dependent Variable: Points scored in game

<em>Since you are attempting to measure how well football players do in different climates, you must measure their scores.</em>

Constants: Uniform, field played in, and ball used

<em>If you use a different uniform, field, or ball, then the data is obviously going to be inconsistent.</em>

Hypothesis: If football players are in colder climates, then they perform better during games.

<em>This is quite self-explanatory: just use the latter data points to craft your hypothesis. Keep in mind that a hypothesis should NEVER be a question!</em>

<em />

I hope I helped! :-)

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A thermos contains 150 cm3 of coffee at 85 ????C. To cool the coffee, you drop two 11-g ice cubes into the thermos. The ice cube

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Answer:

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Explanation:

Heat lost by the coffee = Q

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Density of water = Density of coffee solution = d = 1 g/mL (given)

$Mass =density\times volume$

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Latent heat of ice = $\Delta H_{lat}=334 J/g$

$Q'=334J/g\times 11g=3674 J$

Heat absorbed by the ice after melting = q

Mass of ice melted into water = m' = 11 g

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Initial temperature of water = $T_2$ = 0°C

Final temperature of water = T

$q=m'\times c\times (T_2-T)$

According law of conservation of energy , energy lost by coffee will equal to heat required to melt ice and further to raise the temperature of water.

$-Q=Q'+q$

$-(mc\times (T-T_1))=m'\times c\times (T-T_2)+3674 J$

$150 g\times 4.18 J/g^oC\times (85^oC-T)=11 g\times 4.18 J/g^oC\times (T-0^oC)+3674 J$

On solving we get:

T = 73.73°C

73.73°C is the final temperature of the coffee.

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The given data is:

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