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hichkok12 [17]
1 year ago
15

A glass optical fiber is used to transport a light ray across a long distance. The fiber has an index of refraction of 1.510 and

is submerged in water, which has an index of refraction of 1.333. What is the critical angle (in degrees) for the light ray to remain inside the fiber
Physics
2 answers:
d1i1m1o1n [39]1 year ago
8 0

The critical angle (in degrees) for the light ray to remain inside the fiber is 73.13°.

According to snell's law,

n1sinθi = n2sinθr

n1/n2 = sinθr/sinθi

The critical angle is the angle of incidence at the denser medium when the angle of incidence at the less dense medium is 90°.

This means i=C and r = 90°

The Snell's law formula will become

n1/n2 = sinC/sin90°

n2/n1 = 1/sinC

Where n1 is the refractive index of the less dense medium = 1.473

n2 is the refractive index of the denser medium = 1.540

Substituting the values in the formula,

1.540/1.473 = 1/sinC

1.045 = 1/sinC

SinC = 1/1.045

SinC = 0.957

C = sin^-1(0.957)

C = 73.13°

Learn more about critical angle here brainly.com/question/15009181

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Svetradugi [14.3K]1 year ago
5 0

The answer is 73.13°.

According to snell's law,

n1sinθi = n2sinθr

n1/n2 = sinθr/sinθi

The critical angle is the angle of incidence at the denser medium when the angle of incidence at the less dense medium is 90°

This means i=C and r = 90°

The Snell's law formula will become

n1/n2 = sinC/sin90°

n2/n1 = 1/sinC

Where n1 is the refractive index of the less dense medium = 1.473

n2 is the refractive index of the denser medium = 1.540

Substituting the values in the formula,

1.540/1.473 = 1/sinC

1.045 = 1/sinC

SinC = 1/1.045

SinC = 0.957

C = sin^-1(0.957)

C = 73.13°

The refractive index of glass is 1.5. It way that the velocity of mild in glass is 1.5 times slower than the rate of mild in a vacuum, the speed of light in glass isn't always unbiased of the shade of mild.

Refractive index is likewise the same as the velocity of light c of a given wavelength in an empty area divided with the aid of its speed v in a substance or n = c/v.

The index of refraction of fabric is a ratio that compares the velocity of light in a vacuum ( c=3.00x108ms ) to the velocity of mild in that precise medium. Because the index of refraction increases, the amount that the material bends the mild increases.

Learn more about the index of refraction here brainly.com/question/12469161

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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0
3 years ago
For a relative frequency distribution, relative frequency is computed is computed as ____________.
Yuliya22 [10]

Answer:

For a relative frequency distribution, relative frequency is computed as the class frequency divided by the number of observations.

6 0
2 years ago
a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

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8 0
1 year ago
A 94.7 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.75 r
Marta_Voda [28]

Answer:

The angular velocity of the platform is 1.114 rad/s.

Explanation:

Step 1:  Given data

Mass of the horizontal circular platform = 94.7 kg

Mass of the monkey = 21.1 kg

Initial angular velocity = 1.75 rad/s

A monkey drops a 9.25 kg bunch of bananas

They hit the platform at 4/5 of its radius from the center

Model the platform as a disk of radius 1.63 m

Step 2: Calculate the moment of inertia of the disk

I = ½ * m * r² = ½ * 94.7 * 1.63² = 125.80

Step 3: Calculate the initial angular momentum

I = 125.80 * 1.75 = 220.15

Step 4: Calculate the moment of inertia for the bananas

For the bananas, r = 4/5 * 1.63 = 1.304 m

I = 9.25 * 1.304² = 15.73

Step 5: Calculate Moment of inertia for the monkey

I = 21.1 * 1.63² = 56.06

Step 6: Total moment of inertia = 125.80 + 15.73 + 56.06 = 197.59

Step 7: Calculate final angular momentum = 197.59 * ω

197.59 * ω = 220.15

ω = 220.15 / 197.59

This is approximately 1.114 rad/s.

3 0
3 years ago
Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
Burka [1]

Answer:

1.91773\times 10^{37}\ kg

Explanation:

v = Orbital speed = 130 km/s

d = Diameter = 16 ly

r = Radius = \dfrac{d}{2}=\dfrac{16}{2}=8\ ly

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

1\ ly=9.461\times 10^{15}\ m

As the centripetal force balances the gravitational energy we have the following relation

\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow M=\dfrac{v^2r}{G}\\\Rightarrow M=\dfrac{130000^2\times 8\times 9.461\times 10^{15}}{6.67\times 10^{-11}}\\\Rightarrow M=1.91773\times 10^{37}\ kg

Mass of the the massive object at the center of the Milky Way galaxy is 1.91773\times 10^{37}\ kg

4 0
3 years ago
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