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3 years ago

Light travels through a substance at 1.97 x 10^8 m/s. What is the index of the liquid? What type of liquid is it?

Physics

2 answers:

viktelen [127]3 years ago

6 0

n = 1.52

Index of the liquid = 1.52

The type of the liquid is Crown glass.

Explanation:

In finding the index of refraction, we use this formula

n = c/v

where n is the index of refraction

c is velocity of light in vacuum which is constant and is equivalent to $3 \times 10^{8} \mathrm{m} / \mathrm{s}$

v is the velocity of light in a particular medium

Here we use $1.97 \times 10^{8} \mathrm{m} / \mathrm{s}$

Using the formula,

n = c/v

n = $3 \times 10^{8} / 1.97 \times 10^{8}$

n = 1.52284264

n = 1.52

Thus, the index of the liquid is 1.52

Deffense [45]3 years ago

5 0

The refractive index of the liquid is 1.52. The substance is a crown glass.

Explanation:

Given:

Velocity of light through the substance, v = 1.97 X 10⁸ m/s

Refractive index of the liquid, μ = ?

We know,

μ = $\frac{c}{v}$

where, c = velocity of light in air

v = velocity of light in the substance

speed of light in air is 3 X 10⁸m/s

$= \frac{3 X 10^8}{1.97 X 10^8} \\\\= 1.52\\$

Therefore, refractive index of the liquid is 1.52. The substance is a crown glass.

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Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

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Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

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3 years ago

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Answer:

Here the circuit in which a 4Ω resistor resistor is connected in series and two 8Ω resistor resistors are connected in parallel. Also, ammeter and voltmeter connected in series and parallel circuit respectively.

Now,

The maximum power of each resistance is 16 W

The 4Ω resistor is linked in series with the circuit.

so, P o w e r = I

two

R, here i is the current through the resistor resistor R

1 6 = I

two

∗ 4 Ω

i = 2A

Now 2A passes through parallel resistors of 8Ω resistance.

we know that, in parallel, the potential difference must be constant,

the current is divided into two parts, because the same resistance current in each resistance will be half. then the current through each resistor in parallel is

2 A

two

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So finally the current through the 4Ω resistor = 2 A

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Explanation:

I hope this answer has helped you

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