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3 years ago

Can someone please describe surface tension to me. Please don't give me the dictionary definition. Thank you

Physics

1 answer:

11111nata11111 [884]3 years ago

6 0

Surface tension - My definition -

It's exactly what it says - The tension of a surface with a liquid (such as water), caused by the attraction of the surfaces layer ---- I hope this helps ---- I actually did research it and got some of this from a dictionary, but I changed some of it, too.... Sorry if this doesn't help :)

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A 50 kg boy running with a velocity of 6 m/s passes by a 2000 kg truck at rest. Which one has greater momentum, the truck or the

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Answer:

The boy has a greater momentum

Explanation:

The reason why the boy has a greater momentum is because the truck is at rest so it isn’t moving and momentum has to do with the quantity of motion

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3 years ago

Light traveling in water (n = 1.33) into an unknown medium.If rhe angles if incidence and refraction are 40 degrees and 25 degre

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3 years ago

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denis-greek [22]

Answer:

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Explanation:

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3 years ago

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You have two photos of a person walking. One shows the person at the corner of Third and Main streets, the other shows the perso

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Answer:

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3 years ago

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