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Sedaia [141]
3 years ago
15

a light ray traveling in air enters a second medium and its speed slows to 1.71 × 10^8 meters per second. what is the absolute i

ndex of refraction of the second medium? (1) 1.00 (3) 1.75 (2) 0.570 (4) 1.94
Physics
2 answers:
givi [52]3 years ago
8 0
The absolute refractive index is equal to the speed of light of the wave in air divided by the speed of light in the second medium. This means that it is equal to 3 x10^8 / 1.71 x10^8. This means the answer is 1.75
Dmitriy789 [7]3 years ago
8 0
Absolute Index of Refraction 

                                               =  Velocity of Light in air / Velocity of light in medium

                                               =   3 * 10⁸  m/s   / 1.71 * 10⁸ m/s

                                               = 1.75

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Supposedly there is life on Mars but it’s early to say for sure
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3 years ago
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At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel
sleet_krkn [62]

Answer:

47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

Length of swimming course (L) =..?

The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

L² = 2225

Take the square root of both side.

L = √2225

L = 47.1 ≈ 47 m

Therefore, the length of the swimming course is approximately 47 m.

7 0
3 years ago
A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen
Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

a=\dfrac{v^2}{R}

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

a'=\dfrac{v^2}{R'}

a'=\dfrac{v^2}{R/4}

a'=4\times \dfrac{v^2}{R}

a'=4\times a

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0
3 years ago
Date<br>Page<br>What are the advantages of alcohol<br>thermometric liquid?<br>​
Leya [2.2K]

Answer:

low freezing point. high vapour pressure.

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>WILL</em><em> </em><em>HELP</em><em> </em><em>U</em><em>! </em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em>

3 0
3 years ago
A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla
miss Akunina [59]

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x  10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N

The work done between the ends of the plate is equal to product of the  magnitude of net force on the ball and the distance traveled by the ball.

W = F_{net} *h\\\\W = 15.328 *10^{-3} *  h

W = K.E

15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m

Therefore, the ball traveled 0.827 m

4 0
4 years ago
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