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shtirl [24]
3 years ago
9

Calculate the index of refraction for a medium in which the speed of light is 2.1x 108 m/s. The speed of light in vacuum is 3x10

8 m/s. Your answer should be a number with three decimal places, do not include the unit.
Physics
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

n = 1.42

Explanation:

The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42

So, the refractive index of the medium is 1.42.

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A stone is dropped from from rest at the top of a mine shaft. It takes 95 seconds for the stone to fall to the bottom of the min
Marianna [84]

Distance of fall from rest,
without air resistance              =  (1/2) (gravity) (time)²

                                             = (1/2) (9.8 m/s²) (95 sec)²

                                             =  (4.9 m/s²) (9,025 sec²)

                                             =        44,222.5 meters  .

The depth of the mine shaft is five times the height of Mt. Everest !


6 0
2 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
What is the range of a ball thrown horizontally at 12 m/s if its time of flight is 3.0 s?
vladimir1956 [14]

Answer:

6

Explanation:

because I did this assignment, :) your welcome

Next time do it by yourself, but here's the answer kid

7 0
3 years ago
Read 2 more answers
A 10,000-watt radio station transmits at 535 kHz. Determine the number of joules transmitted per second. 10,000 J/s 10 J/s 535 J
REY [17]
1 nanowatt  =  1 nanojoule/sec
1 watt  =  1 joule/sec
10 watts  =  10 joules/sec
100 watts  =  100 joules/sec
742.914 watts  =  742.914 joules/sec
1,000 watts  =  1,000 joules/sec
10,000 watts  =  10,000 joules/sec
100,000 watts  =  100,000 joules/sec
1 megawatt  =  1 megajoule/sec
1 gigawatt  =  1 gigajoule/sec
1 petawatt  =  1 petajoule/sec

We don't care what frequency the transmission is using,
or who their morning DJ is.

5 0
3 years ago
You are driving a 2400.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an inters
olya-2409 [2.1K]

Answer:0.43

Explanation:

Given

mass of car m=2400 kg

Speed of car u=14 m/s

Distance traveled before coming to halt s=23.2 m

Let \muthe coefficient of friction

Maximum deceleration road can provide during motion is

a=\mu g

using v^2-u^2=2 as

0-14^2=2\cdot (-\mu g)\cdot 23.2

\mu =\frac{196}{454.72}

\mu =0.431

7 0
2 years ago
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