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3 years ago

If the index of refraction of a medium is 1.4, determine the speed of light in that medium.

Physics

1 answer:

Damm [24]3 years ago

3 0

The speed of light in that medium is $2.14 \times 10^8 \ m/s$ .

<u>Explanation:</u>

It is known that the light's speed is constant when it travels in vacuum and the value is $3 \times 108 m/s$ . When the light enters another medium other than vacuum, its speed get decreased as the light gets refracted by an angle.

The amount of refraction can be determined by the index of refraction or refractive index of the medium. The refraction index is measured as the ratios of speed of light in vacuum to that in the medium. It is represented as η = $\frac {c}{v}$

So, here η is the index of refraction of a medium which is given as 1.4, c is the light's speed in vacuum ( $3 \times 10^8 ms^-^1$ ) and v is the light's speed in that medium which we need to find.

$1.4= \frac{(3 \times 10 ^ 8)} {v}$

$v= \frac {(3 \times 10^8)}{1.4} =2.14 \times 10^8 \ m/s$

Thus the speed of light in that medium is $2.14 \times 10^8 \ m/s.$

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A 20 Ohm resistance is connected

frutty [35]

Answer: 225 V

Explanation:

<u>Given:</u>

Secondary voltage,V2 = 150v

Resistance is connected across secondary winding,∴R2 = 20Ω

Supply current, ie, I1 = 5A

$\text{Now, secondary current,} \ $I_{2}=\frac{V_{2}}{R_{2}}=\frac{150}{20}=7.5$$$\therefore I_{2}=7.5 \mathrm{~A}$$For Transformers, we have $\frac{N_{2}}{N_{1}}=\frac{V_{2}}{V_{1}}=\frac{I_{1}}{I_{2}}$ so, Taking $\frac{N_{2}}{N_{1}}=\frac{l_{1}}{l_{2}}$ inserting all given \& obtained values, we get $\frac{N_{2}}{N_{1}}=\frac{5}{7.5}$$$\therefore \text { Turns ratio }=\frac{N_{1}}{N_{2}}=\frac{7.5}{5}=\frac{3}{2}=3: 2$$$

$\text{Again taking,}\frac{N_{1}}{N_{2}}=\frac{V_{2}}{V_{1}}$ \\So, \\$V_{1}=\frac{N_{1}}{N_{2}} V_{2}\\$ inserting all values, we get, $V_{1}=\frac{7.5}{5} * 150=225$\\Therefore, \fbox{Primary voltage, {$V_{1}=225 \mathrm{~V}$}}$

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2 years ago

Explain why friction makes it harder to push a box filled with groceries than an empty box

kupik [55]

It is harder to push a box with groceries in it because the mass is weighing the box down. Which makes it harder to push

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3 years ago

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

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2 years ago

Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the c

AleksAgata [21]

Answer:

*If the particles are deflected in opposite directions, it implies that their charges must be opposite

*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

Explanation:

When a charged particle enters a magnetic field, it is subjected to a force given by

F = q v x B

where bold letters indicate vectors

this expression can be written in the form of a module

F = qv B sin θ

and the direction of the force is given by the right-hand rule.

In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1

If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.

Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

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Why are there temperature differences on the moon's surface even though there is no atmosphere present?

nika2105 [10]

The lack of an atmosphere means convection cannot happen on the moon. Therefore, there is no form of heat dissipation on regions in direct sunlight. In addition, the lack of an atmosphere means there is no greenhouse effect on the moon. This is why regions facing away from sunlight are very cold.

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3 years ago