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4 years ago

15

A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i

n the material?

2 answers:

Len [333]4 years ago

7 0

Answer:

The percentage of speed of light in vacuum to the speed of light in the said material is 80%

Explanation:

The common values of refractive index are between 1 and 2 since nothing can travel faster than the speed of light, therefore, no material has a refractive index lower than 1.

According to the formula $n=\frac{c}{v}$

where $n$ is the index of refraction

$c$ is the speed of light in vacuum

and $v$ is the speed of light in the material, it can be seen that n and v are inversely proportional which means greater the refractive index lower is the speed of light.

Since we know that speed of light in vacuum is 300,000 km/s using the formula we get,

$v=\frac{c}n}$

$v=\frac{300000}{1.25}=240,000 km/s$

for finding percentage,

$=\frac{240000}{300000}*100 = 80$ %

beks73 [17]4 years ago

3 0

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

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Answer:

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2 years ago

0.054x2.33x90............

guapka [62]

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3 years ago

Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to

stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

$\mu_{sB}=0.126$

$\mu_{sC}=0.168$

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

$\sum \tau_{A}=0$

$N(3m)-W(1.5m)=0$

When solving for N we get:

$N=\frac{W(1.5m)}{3m}$

$N=\frac{(1962N)(1.5m)}{3m}$

$N=981N$

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

$\sum F_{y}=0$

$-F_{By}+N_{c}=0$

$F_{By}=N_{c}$

Next, the forces in x.

$\sum F_{x}=0$

$-f_{sB}-f_{sC}+P_{x}=0$

We can find the x-component of force P like this:

$P_{x}=360N(\frac{4}{5})=288N$

and finally the torques about C.

$\sum \tau_{C}=0$

$f_{sB}(1.75m)-P_{x}(0.75m)=0$

$f_{sB}=\frac{288N(0.75m)}{1.75m}$

$f_{sB}=123.43N$

With the static friction force in point B we can find the coefficient of static friction in B:

$\mu_{sB}=\frac{f_{sB}}{N}$

$\mu_{sB}=\frac{123.43N}{981N}$

$\mu_{sB}=0.126$

And now we can find the friction force in C.

$f_{sC}=P_{x}-f_{xB}$

$f_{sC}=288N-123.43N=164.57N$

$f_{sC}=N_{c}\mu_{sC}$

and now we can use this to find static friction coefficient in point C.

$\mu_{sC}=\frac{f_{sC}}{N}$

$\mu_{sC}=\frac{164.57N}{981N}$

$\mu_{sB}=0.168$

3 0

3 years ago

1. What is the wave speed of a wave that has a frequency of 100 Hz and a wavelength of 0.30 m?

aivan3 [116]

Answer:

1. v = 30 m/s

2. v = 5 m/s

3. f = 40 Hz

4. f = 400 Hz

5. f = 300 Hz

6. λ = 0.772 m

7. λ = 0.386 m

8. λ = 0.625 m

9. v = 100 m/s

10. v = 50 m/s

Explanation:

The relationship between frequency, wavelength, and speed of a wave is given by the following formula:

$v = f\lambda$

where,

v = speed of wave

f = frequency of wave

λ = wavelength

1.

f = 100 Hz

λ = 0.3 m

Therefore,

v = (100 Hz)(0.3 m)

<u>v = 30 m/s</u>

<u></u>

2.

f = 50 Hz

λ = 0.1 m

v = (50 Hz)(0.1 m)

<u>v = 5 m/s</u>

<u></u>

3.

v = 20 m/s

λ = 0.5 m

$f = \frac{v}{\lambda} = \frac{20\ m/s}{0.5\ m}$

<u>f = 40 Hz</u>

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4.

v = 80 m/s

λ = 0.2 m

$f = \frac{v}{\lambda}=\frac{80\ m/s}{0.2\ m}$

<u>f = 400 Hz</u>

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5.

v = 120 m/s

λ = 0.4 m

$f = \frac{v}{\lambda}=\frac{120\ m/s}{0.4\ m}$

<u>f = 300 Hz</u>

<u></u>

6.

v = 340 m/s

f = 440 Hz

$\lambda = \frac{v}{f}=\frac{340\ m/s}{440\ Hz}\\$

<u>λ = 0.772 m</u>

<u></u>

7.

v = 340 m/s

f = 880 Hz

$\lambda = \frac{v}{f}=\frac{340\ m/s}{880\ Hz}\\$

<u>λ = 0.386 m</u>

<u></u>

<u></u>

8.

v = 250 m/s

f = 400 Hz

$\lambda = \frac{v}{f}=\frac{250\ m/s}{400\ Hz}\\$

<u>λ = 0.625 m</u>

<u></u>

9.

f = 50 Hz

λ = 2 m

v = (50 Hz)(2 m)

<u>v = 100 m/s</u>

<u></u>

10.

f = 100 Hz

λ = 0.5 m

v = (100 Hz)(0.5 m)

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6 0

3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

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3 years ago