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2 years ago

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum. Find the refractive index of the medium.

Physics

2 answers:

nalin [4]2 years ago

8 0

Refractive index is defined as

$\mu = \frac{speed of light in air}{speed of light in medium}$

$\mu = \frac{c}{v}$

given that

v = 0.6 c

so we will have

$\mu = \frac{c}{0.6c}$

$\mu = 1.67$

grin007 [14]2 years ago

3 0

<h2>Answer:</h2>

The refractive index is 1.66

<h2>Explanation:</h2>

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum .

Refractive index of medium = speed of light in vacuum / speed of light in medium

RI = 1/0.6 = 5/3 or 1.66

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A car travels north along a straight highway at an average speed of 85 km/h. After driving 2.0 km, the car passes a gas station

Sati [7]

Answer:

x_total = 23250 m

Explanation:

This is a uniform motion exercise, the equation that describes the motion is

v = x / t

the position is

x = v t

let's reduce the magnitude to the SI system

v = 85 km / h (1000 m / 1km) (1h / 3600 s) = 23.61 m / s

t = 0.25 h (3600 s / 1h) = 900 s

let's calculate

x₁ = v t₁

x₁ = 23.61 900

x₁ = 21250 m

The distance from the origin is

x_total = x_station + x₁

x_total = 2000 + 21250

x_total = 23250 m

5 0

2 years ago

Calculate the amount of energy produced in a nuclear reaction in which the mass defect is 0.187456 amu.

luda_lava [24]

For nuclear reactions, we determine the energy dissipated from the process from the Theory of relativity wherein energy is equal to the mass defect times the speed of light. We calculate as follows:

E = mc^2 = 0.187456 (3x10^8)^2 = 1.687x10^16 J

Hope this answers the question.

8 0

2 years ago

Read 2 more answers

A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t

Pie

Answer:

$I = \frac{mvb}{6}$

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

$L = I$ ω .... (1)

moment of inertia of rod rotating about its center of length b

$I = \frac{ mb^2}{12}$ ........ .(2)

using v = ωr

where w is angular velocity

and r is radius of rod which is equal to b

so we get 2v = ωb

ω = 2v/b ................. (3)

here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

$L = \frac{mb^2}{12}$ × $\frac{2v}{b}$

so final answer will be $L = \frac{mvb}{6}$

7 0

2 years ago

At 213.1 K a substance has a vapor pressure of 45.77 mmHg. At 243.7 K it has a vapor pressure of 193.1 mm Hg. Calculate its heat

vlabodo [156]

Answer:

$20.3125$ kJ/mol

Explanation:

$P_{i}$ = initial vapor pressure = 45.77 mm Hg

$P_{f}$ = final vapor pressure = 193.1 mm Hg

$T_{i}$ = initial temperature = 213.1 K

$T_{f}$ = final temperature = 243.7 K

$H$ = Heat of vaporization

Using the equation

$ln\left ( \frac{P_{f}}{P_{i}} \right ) = \left ( \frac{-H}{R} \right )\left ( \frac{1}{T_{f}} - \frac{1}{T_{i}}\right)$

$ln\left ( \frac{193.1}{45.77} \right ) = \left ( \frac{-H}{8.314} \right )\left ( \frac{1}{243.7} - \frac{1}{213.1}\right)$

$H = 20312.5$ J/mol

$H = 20.3125$ kJ/mol

8 0

2 years ago

Un tren emplea cierto tiempo en recorrer 240 km. Si la velocidad hubiera sido 20 km por hora mas que la que llevaba hubiera tard

podryga [215]

Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x•a

x = 240 / a Equation 1

For stage two

d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

Substitute equation 1 into 2

240 = (240/a + 20) × (a -2)

240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

Then, t_2 = 6 - 2 = 4 hours

t_1 = 6 hours

t_2 = 4 hours

Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

5 0

2 years ago