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stealth61 [152]

2 years ago

14

A glass optical fiber is used to transport a light ray across a long distance. The fiber has an index of refraction of 1.540 and

is submerged in glycerin, which has an index of refraction of 1.473. What is the critical angle (in degrees) for the light ray to remain inside the fiber

1 answer:

Zinaida [17]2 years ago

4 0

Answer:

73.13°

Explanation:

According to snell's law,

n1sinθi = n2sinθr

n1/n2 = sinθr/sinθi

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°

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An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

ivolga24 [154]

1) Focal length

We can find the focal length of the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$ (1)
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

In this case, $d_o = 18 cm$ , while $d_i=-4 cm$ (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
$\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}$
from which we find
$f=- \frac{36}{7} cm=-5.1 cm$

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
$r=2f=2 \cdot 5.1 cm=10.2 cm$

3 0

2 years ago

In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What

allsm [11]

Answer:

Force constant, k = 653.3 N/m

Explanation:

It is given that,

Weight of the bag of oranges on a scale, W = 22.3 N

Let m is the mass of the bag of oranges,

$m=\dfrac{W}{g}$

$m=\dfrac{22.3}{9.8}$

m = 2.27 kg

Frequency of the oscillation of the scale, f = 2.7 Hz

We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$k=4\pi^2 f^2m$

$k=4\pi^2 \times (2.7)^2\times 2.27$

k = 653.3 N/m

So, the force constant of the spring of the scale is 653.3 N/m. Hence, this is the required solution.

7 0

3 years ago

your drop a coin from the top of a hundred-story building(1000m). If you ignore air resistance, how fast will it be falling righ

Ede4ka [16]

consider the motion of con from top to bottom

Y = vertical displacement = 1000 m

a = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity at the top = 0 m/s

v = final velocity at the bottom = ?

using the kinematics equation

v² = v²₀ + 2 aY

v² = 0² + 2 (9.8) (1000)

v = 140 m/s

t = time taken to hit the ground

Using the equation

v = v₀ + at

140 = 0 + 9.8 t

t = 14.3 sec

7 0

2 years ago

Read 2 more answers

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

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2 years ago

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Lunna [17]

Answer:

Because the chemical energy is converted to electrical enegy in the cell phone

Explanation:

This set up is not a voilation of the first law of thermodynamics because the chemical energy is not lost, it is only being transformed into electrical energy.

According to the first law of thermodynamics "energy is neither created nor destroyed but transformed from one form to another".

In a charged cell phone, the chemical energy is being transformed to electrical energy as you use it.
A phone cannot directly be powered with chemical energy.
The energy transformation to electrical energy makes it possible to use the phone.
In this process no energy is lost nor created. But simply, energy is transformed.

6 0

2 years ago