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3 years ago

18. A swimmer steps out of a swimming pool into a light breeze. Why does

1 answer:

OverLord2011 [107]3 years ago

6 0

When the swimmer gets out, the water on her skin begins to evaporate. This process requires takes a lot of energy and that energy comes from the heat of her body. This is similar to when someone sweats. Hope this helps!

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Calcula la densidad del hierro , si 393 g ocupan un volumen de 50 ml

eimsori [14]

Density is mass over volume: d=m/v
d = 393/50 => d = 7.86 g/ml

7 0

3 years ago

An object is placed a distance of twice the focal length away from a diverging lens. What is the magnification of the image?

Lesechka [4]

Answer:

+1/3

Explanation:

The lens equation states that:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

For a diverging lens, the focal length is negative: $f=-f$

and we also know that the object is placed a distance of twice the focal length, so $p=2f$

So we can find q from the equation above

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-f}-\frac{1}{2f}=-\frac{3}{2f}\\q=-\frac{2}{3}f$

And the magnification of the image is given by

$M=-\frac{q}{p}=-\frac{-\frac{2}{3}f}{2f}=\frac{1}{3}$

5 0

4 years ago

Read 2 more answers

Which example represents the impact of religion on culture in the United States? A. children receiving counseling at school B. f

dmitriy555 [2]

Answer:

D. Banks and offices closing on sunday.

Explanation:

Church is usually on sunday and people need a day off to do it because service is about 3 hours long. When banks and offices close on this day it shows how religion has affected our lives and our culture.

5 0

3 years ago

A student dips a strip of metal into a liquid. Which is evidence that only a physical change has occurred?

sattari [20]

Answer:

PLS add the picture too.....

3 0

3 years ago

A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

zzz [600]

Answer:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Explanation:

We know that the electric force equation is:

$F=k\frac{q_{1}*q_{2}}{r^{2}}$

k is the electric constant $9*10^{9} Nm^{2}/C^{2}$
r is the distance between the particles
q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

$F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}$

$F_{31}=k\frac{q*2q}{r_{31}^{2}}$

$F_{31}=k\frac{2q^{2}}{r_{31}^{2}}$

r(31) is the distance between 3 and 1

2. Now, let's find the electric force between the third particle and the second particle.

$F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}$

$F_{32}=k\frac{q*(-q)}{r_{32}^{2}}$

$F_{32}=-k\frac{q^{2}}{r_{32}^{2}}$

r(32) is the distance between 3 and 2.

Now, $r_{31}+r_{32}=2a$ or $r_{32}=2a-r_{31}$

The net force must be zero so:

$F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}$

We just need to solve it for r(31)

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Therefore the distance from the origin will be:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

I hope it helps you!

4 0

4 years ago