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3 years ago

This is actually physical science but there isn’t a tag for that☹︎

Physics

1 answer:

gayaneshka [121]3 years ago

3 0

500ml = 0.5 L
4.2L = 4200ml
400mg = 0.0004kg
150kg = 150000g

I’m not sure about number 4 though

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A badger is running at a speed of 1 m/s. If the badger moves that was for 2600 seconds, how far will the badger travel?

katen-ka-za [31]

Answer:

It would be 2600

Explanation:

M/S stands for meters per second. If it moved 1 meter for 2600 seconds, than it would be 2600. You just multiply 2600 by 1! I hope this helps :D

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3 years ago

A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l

son4ous [18]

Answer:

The focal length of the lens in ethyl alcohol is 41.07 cm.

Explanation:

Given that,

Refractive index of glass= 1.500

Refractive index of air= 1.000

Refractive index of ethyl alcohol = 1.360

Focal length = 11.5 cm

We need to calculate the focal length of the lens in ethyl alcohol

Using formula of focal length for glass air system

$\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Using formula of focal length for glass ethyl alcohol system

$\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Divided equation (II) by (I)

$\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}$

Where, $n_{g}$ = refractive index of glass

$n_{a}$ = refractive index of air

$n_{ethyl}$ = refractive index of ethyl

Put the value into the formula

$\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}$

$\dfrac{f'}{11.5}=\dfrac{25}{7}$

$f'=\dfrac{25}{7}\times11.5$

$f'=41.07\ cm$

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

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3 years ago

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3 years ago

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gladu [14]

Answer:

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Explanation:

because i just took it and got it correct.

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