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2 years ago

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Mike understands the formula for temperature conversion, but is having difficulty imagining what various temperatures feel like.

which of these would be a good example to tell mike what 37°c feels like? a. room temperature b. a hot pan c. body temperature d. a chilly autumn evening

1 answer:

Alisiya [41]2 years ago

3 0

A good example that will help to to tell mike what 37°c feels like is body temperature.

<h3>What is room temprrature?</h3>

The average room temperature is typically around 20°C, or 68 degrees Fahrenheit.

<h3>What is body temperature?</h3>

The average body temperature is 98.6 Fahrenheit or 37 degrees Celsius.

Thus, a good example that will help to to tell mike what 37°c feels like is body temperature.

Learn more about body temperature here: brainly.com/question/5295345

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sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

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