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4 years ago

9

If only one external force acts on a particle, does it necessarily change the particle's kinetic energy and velocity? (select al

l that apply.)

1 answer:

ivann1987 [24]4 years ago

7 0

It's necessary to change the particles kinetic and velocity to open external and eternal.

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Examine the weather map.

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Answer:

High pressure.

Explanation:

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Mike has been struggling to maintain a healthy weight since starting college.which of the following statements shows a positive

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Tropospheric ozone ________.

Nata [24]

Tropospheric ozone is produced through interaction of heat and UV light, with nitrogen oxides and carbon containing compounds.

Option: D

Explanation:

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3 years ago

What term represents the total length of a path taken by an object?

denpristay [2]

Hey there!

I would say either B or C

Hope this helps you!

God bless ❤️

xXxGolferGirlxXx

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3 years ago

A light on the ground is 30 feet away from a building. A 5 foot tall man is walking from the light to the building at a rate of

mars1129 [50]

Answer:

the shadow is shrinking at 4.5ft/s

Explanation:

Calculing the rate change is a problem that evolves derivates, so you have to find a function that represent the height of the shadow and then derivate that with respect to time.

So first we have to find the function that express the height of the shadow, to do that notice that both the green triangles have the same common blue angle that i pictured in the image, so if we express the tangent function for that angle we have:

$Tan (\alpha)=\frac{5ft(height of the person)}{x(distance of the person from the light)}=\frac{H(Height of the shadow)}{30ft(distance between the ligth and the building)}$

and from here we can tell that:

$H=\frac{5ft*30ft}{x}=\frac{150ft}{x}$

now that we have our function we derivate with respect to time, doing that we have:

$\frac{dH}{dt}=\frac{d}{dt}(\frac{150ft^{2} }{x} )=\frac{-150ft^{2}}{x^{2}}*\frac{dx}{dt}$

where $\frac{dx}{dt}= 3ft/s$

now we just have to replace 10 ft in x and that is the changing rate of the shadow height:

$\frac{dH}{dt}(10ft)=\frac{-150ft^{2}}{(10ft)^{2}}*\frac{3ft}{s}=-4.5ft/s$

and that is it, the minus sign is because the shadow is shrinking instead of expanding.

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3 years ago