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2 years ago

Process of changing a liquid into a gas

Physics

2 answers:

vazorg [7]2 years ago

5 0

Boiling. The temperature at which a liquid begins to boil is called the <span>boiling point.</span>

Otrada [13]2 years ago

5 0

Answer:

Evaporation jsut took teST LOL

Explanation:

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2 Polnis

Marta_Voda [28]

Answer:

Oprion A

The length of time over which the conditions are measured

Explanation:

Weather captures the daily atmospheric conditions over a short duration hence it is short-term. Climate is the average of weather and covers a longer duration hence long-term. Therefore, what differentiatea these two is the length of time over which the conditions are measured.

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3 years ago

For the element 35 17 C1, give the number of... 1. protons 2. neutrons 3. electrons

faltersainse [42]

Answer:

Chlorine 35:

Atomic mass 35.43

Number of protons 17

Number of electrons: 17

Number of neutrons: 18

Explanation:

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3 years ago

The amp is the unit for _________.

adell [148]

B.) <span>The amp is the unit for "Current"

Hope this helps!</span>

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2 years ago

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Please help, it's super easy!

Alexandra [31]

The moon's gravitational pull on Earth causes water to bulge on two sides of the Earth(#3)

https://scijinks.gov/tides/

5 0

2 years ago

Read 2 more answers

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago