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4 years ago

10

you are running with your friend on a hot day. your body begins to warm up. describe how your skin responds to the excess heat.

2 answers:

scoundrel [369]4 years ago

8 0

Your body sweats to to make sure your body doesn't over heat and it regulates your body temperature.

Serga [27]4 years ago

6 0

You start to sweat, its the bodies way of cooling you down.
Hope this helps! :)

Please give brainliest :D

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A man pulls on his dogs leash to keep him from running after the bicycle. Which term best describes this example?

Luda [366]

What term do you mean? like what he did to the dog is he stopped the dog

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3 years ago

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A tourist averaged 82km/hr for a 6.5h trip in her volkswagen. how far did she go?

Whitepunk [10]

82 ÷ 6.5 = 12.615384615384... repeating
round = 12.6
12.6 · 6.5 = 81.9
round = 82
She went an average of 12.6 km an hour
Hope this helps! ;)

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3 years ago

How to convert 45.77 cubic inches to cubic centimeters

Paha777 [63]

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3 years ago

An engine has an imput of 100 j and gives out 27 j of useful kenetic energy what is the efficiency of the engine

Lunna [17]

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Based on what we have learned so far, in what ways have the experiences of Asian-Canadians stayed the same over time? Why do you think these similarities exist? Please give specific examples to support your ideas.

Explanation:

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3 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

(c) 2.86s

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

<u>The answer for (a) is 1.6m</u>

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

<u>the answer to (b) is 6.86m</u>

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

and since $g=9.81m/s^2$

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

At $t=1.43s$ the car hits the water, so the horizontal distance at that time is:

$x(t)=x_{0}+v_{0}t$

$x(1.43)=(2m/s)(1.43s)=2.86m$

<u>the answer to (c) is 2.86s</u>

6 0

3 years ago