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monitta
3 years ago
12

What is the speed of light with N=1.33

Physics
2 answers:
alexgriva [62]3 years ago
3 0

If the refractive index of some substance is  1.33, then
the speed of light in that substance is
                  
               (speed of light in vacuum) / (1.33)  = 

                         (299,792,458 m/s) / (1.33)  =  <em>225,407,863 m/s</em>


liq [111]3 years ago
3 0
v=2.25` 10^{8}  \frac{m}{s}
N=1.33 being the Index of Refraction of a material and c=3.0` 10^{8}  \frac{m}{s} the speed of light in vacuum you get:
v= \frac{c}{N}= \frac{3.0 10^{8} }{1.33}=2.25` 10^{8}  \frac{m}{s} speed of light in that material.
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a mars prototype carrying scientific instruments has a mass of 1060kg. a net force of 52000 n is applied to this roverat a test
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F = 52000 N

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6 0
3 years ago
An amoeba has 1.00 x 1016 protons and a net charge of 0.300 pC. Assuming there are 1.88 x 106 fewer electrons than protons, If y
Xelga [282]

Answer:

The fraction of the protons would have no electrons =1.88\times 10^{-10}

Explanation:

We are given that

Amoeba has total number of protons=1.00\times 10^{16}

Net charge, Q=0.300pC

Electrons are fewer than protons=1.88\times 10^6

We have to find the fraction of protons would have no electrons.

The fraction of the protons would have no electrons

=\frac{Fewer\;electrons}{Total\;protons}

The fraction of the protons would have no electrons

=\frac{1.88\times 10^{6}}{1.00\times 10^{16}}

=1.88\times 10^{-10}

Hence, the fraction of the protons would have no electrons =1.88\times 10^{-10}

6 0
3 years ago
What is the closed physical system?
Alexus [3.1K]
Answer:D
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8 0
3 years ago
1) Displacement is to velocity as ____________ is to acceleration. A) direction B) speed C) time D) velocity
vivado [14]

Answer: C) time

Explanation:

7 0
3 years ago
A spotlight on a boat is y = 2.2 m above the water, and the light strikes the water at a point that is x = 8.5 m horizontally di
slava [35]

Answer:

The answer to the question is

The distance d, which locates the point where the light strikes the bottom is   29.345 m from the spotlight.

Explanation:

To solve the question we note that Snell's law states that

The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction

n₁sinθ₁ = n₂sinθ₂

y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and

θ₁ =  14.511 °

n₁ = 1.0003 = refractive index of air

n₂ = 1.33 = refractive index of water

Therefore sinθ₂ =  \frac{n_1sin\theta_1}{n_2}  = \frac{1.003*0.251}{1.33} = 0.1885 and θ₂ = 10.86 °

Since the water depth is 4.0 m we have tanθ₂ = \frac{4}{x_2} or x₂ = \frac{4}{tan\theta_2 } =\frac{4}{tan(10.86)} = 20.845 m

d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.

5 0
3 years ago
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