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2 years ago

What is speed of light in crown glass?

Physics

1 answer:

german2 years ago

5 0

Answer:

1.97 x 10^8 m/s

Explanation:

refractive index of crown glass with respect to air, n = 1.52

speed of light in air, c = 3 x 10^8 m/s

Let v be the speed of light in crown glass.

By use of the definition of refractive index

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

where, n be the refractive index of crown glass, c be the speed of light in vacuum and v be the speed of light in crown glass

$1.52 = \frac{3 \times 10^{8}}{v}$

$v = \frac{3 \times 10^{8}}{1.52}$

v = 1.97 x 10^8 m/s

Thus, the speed of light in crown glass is 1.97 x 10^8 m/s.

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Describe the kinetic molecular theory

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The kinetic theory of gases is a simple, historically significant model of the thermodynamic behavior of gases, with which many principal concepts of thermodynamics were established. The model describes a gas as a large number of identical submicroscopic particles, all of which are in constant, rapid, random motion

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3 years ago

A 2000kg suv accelerates from rest at a rate of 3.00m/s^2. The total amount of force resisting its motion 1500N. How much force

choli [55]

The total force that the SUV exerts is:

F = 2000 kg * 3 m/s^2

F = 6000 N

Since a resisting force amounting to 1500 N is exerted, then the force exerted by the SUV tires is:

F tire = 6000 N – 1500 N

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2 years ago

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

Lerok [7]

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 = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 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:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

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

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

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2 years ago

A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec

olga2289 [7]

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

$h=u't+\dfrac{1}{2}gt^2$

$h=\dfrac{1}{2}gt^2$

$h=\dfrac{1}{2}\times 9.8\times (4.3)^2$

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

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2 years ago

a spaceship, at rest in a certain reference frame s, is given a speed increment of 0.500c. it is then given a further 0.500c inc

bezimeni [28]

In order to give a spaceship at rest in a specific reference frame s a speed increment of 0.500c, seven increments are required. Then, in this new frame, it receives an additional 0.500c increment.

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Likewise, V3 = 0.929c

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V7 = 0.999c

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6 months ago