Find the speed of light in carbon tetrachlorideethyl alcohol. The refraction index is 1.461 using 3 x 10^8 m/s as the speed of light in vacuum. Answer in units of m/s.

Answers

Answer 1
Answer:

Answer:

2.05 x 10^8 m /s

Explanation:

c = 3 x 10^8 m/s

μ = c / v

where, μ is the refractive index, c be the velocity of light in air and v be the velocity of light in the medium.

μ = 1.461

1.461 = 3 x 10^8 / v

v = 3 x 10^8 / 1.461

v = 2.05 x 10^8 m /s


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Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Answers

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = (1)/(f)

where

f = focal length

Thus

f = (1)/(P)

f = (1)/(2) = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

(1)/(f) = (1)/(u) + (1)/(u')

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

(1)/(u') = (1)/(f) - (1)/(u)

(1)/(u') = (1)/(0.5) - (1)/(0.25)

(1)/(u') = (1)/(f) - (1)/(u)

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.

Answers

Answer:

Explanation:

For a convex mirror, the value of its image distance and its focal length are negative.

using the mirror formula 1/f = 1/u+1/v

f is the focal length = Radius of curvature/2 = 0.560/2

f= 0.28m

u is the object distance = 10.6m

v is the position of the image = ?

On substitution;

1/0.28 = 1/10.6 + 1/-v

3.57 = 0.094 - 1/v

3.57 - 0.094 = -1/v

3.476 = -1/v

v = -1/3.476

v = -0.2877m

B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.

C) Magnification = Image distance/object distance

Magnification  = 0.2877/10.6

Magnification = 0.0271

The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7 T · N/A and the speed of light is 2.99792 × 108 m/s. Find the magnitude of Em for the electromagnetic waves at the top of the atmosphere. Answer in units of N/C.

Answers

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed  ,E=Electric filed

Power  P = I A

A=Area=4πr²  ,I=Intensity

I=(CB^2)/(2\mu_0)

I=(CE^2)/(2\mu_0 C^2)

E=\sqrt{{2I\mu_0 C}}

E=\sqrt{{2* 1150* 4\pi * 10^(-7)(2.99792* 10^8)}}

E=930.84 N/C

Therefore answer is 930.84 N/C

Final answer:

To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.

Explanation:

To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.

Subbing in the given values, we can compute Em as:

Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]

This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.

Learn more about Electromagnetic Waves here:

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With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 41 m

Answers

Answer:

The speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Explanation:

Given;

maximum vertical height of the throw, H = 41 m

Apply the following kinematic equation;

V² = U² + 2gH

where;

V is the final speed with which the ball will rise to a maximum height

U is the initial speed of the ball = 0

g is acceleration due to gravity = 0

V² = U² + 2gH

V² = 0² + 2gH

V² =  2gH

V = √2gH

V = √(2 x 9.8 x 41)

V = 28.35 m/s

Therefore, the speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

A stone is thrown with an initial speed of 11.5 m/s at an angle of 50.0 above the horizontal from the top of a 30.0-m-tall building. Assume air resistance is negligible, and g = 9.8 m/s2. What is the magnitude of the horizontal displacement of the rock?

Answers

Answer:

The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Explanation:

Given that,

Initial speed = 11.5 m/s

Angle = 50.0

Height = 30.0 m

We need to calculate the horizontal displacement of the rock

Using formula of horizontal component

v_(x)=u\cos\theta

Put the value into the formula

v_(x)=11.5*\cos50

v_(x)=7.39\ m/s

Hence, The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Final answer:

The question is about determining the horizontal displacement of a projectile based on the given initial speed and projection angle and the height of the launch. This can be calculated using the equations of motion, specifically those pertaining to projectile motion.

Explanation:

In this problem, we're dealing with projectile motion. The stone being thrown is the projectile in this case. The horizontal displacement, also known as range, of a projectile can be defined using the formula: range = (initial speed * time of flight) * cosθ, where θ is the angle of projection. The initial speed is given as 11.5 m/s and the angle as 50 degrees. Now, we need to calculate the time of flight. This can be found by the formula: time of flight = (2 * initial speed * sinθ) / g. Considering g, the acceleration due to gravity, as 9.8 m/s², we can find the time of flight and thus calculate the range. Always remember that while the vertical motion of a projectile is affected by gravity, the horizontal motion remains constant.

Learn more about Projectile Motion here:

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What does a planet need in order to retain an atmosphere? How does an atmosphere affect the surface of a planet and the ability of life to exist?

Answers

Answer:

Explained

Explanation:

In order to retain atmosphere a planet needs to have gravity. A gravity sufficient enough to create a dense atmosphere around it, so that it can retain heat coming from sun. Mars has shallow atmosphere as its gravity is only 40% of the Earth's gravity. Venus is somewhat similar to Earth but due to green house effect its temperature is very high. Atmosphere has a huge impact on the planets ability to sustain life. Presence of certain kind gases make the atmosphere poisnous for life. The atmosphere should be such that it allows water to remain in liquid form and maintain an optimum temperature suitable for life.