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Papessa [141]
4 years ago
12

An Argon laser (λ = 5.0×102nm) shines down a silica glass fiber-optic cable with index of refraction 1.46. What is the speed of

the laser light, cf , in the cable? Select One of the Following:
(a) 1.5 × 108 m s
(b) 2.1 × 108 m s
(c) 3.0 × 108 m s
(d) 4.4 × 108 m
Physics
2 answers:
kramer4 years ago
8 0

Answer:

2.1 x 10^8 m/s option (b)

Explanation:

refractive index of the fiber, n = 1.46

refractive index of air = 1

According to the definition of refractive index

Speed of light in vacuum / speed of light in fiber = n

speed of light in fiber = speed of light in air / n

                                   = ( 3 x 10^8) / 1.46 = 2.1 x 10^8 m/s

Thus, the speed of light in fiber is 2.1 x 10^8 m/s.

Yanka [14]4 years ago
4 0

Answer:

The speed of the laser light in the cable, c_f=2.1\times 10^8\ m/s

Explanation:

It is given that,

Wavelength of Argon laser, \lambda=5\times 10^2\ nm=5\times 10^{-7}\ m

Refractive index, n = 1.46

Let c_f is the speed of the laser light in the cable. The speed of light in a medium is given by :

c_f=\dfrac{c}{n}

c_f=\dfrac{3\times 10^8\ m/s}{1.46}

c_f=2.05\times 10^8\ m/s

or

c_f=2.1\times 10^8\ m/s

So, the speed of the laser light is 2.1\times 10^8\ m/s. Hence, this is the required solution.

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If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
wel

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

3 0
3 years ago
draw a velocity graph with Vi = 4 m/s and decreasing uniformly so that velocity at 2 seconds is 2 m/s and remaining constant fro
Shkiper50 [21]

For the velocity graph: start at 0s and 4m/s and draw a straight line to 2s and 2 m/s. Then draw a straight horizontal line to 4s and 2m/s

For the acceleration graph: start with a horizontal line from 0s and 2m/s/s to 2s and 2m/s/s. The draw another line from 2 s and 0m/s/s to 4 s and 0m/s/s

6 0
3 years ago
When you whirl a can at the end of a string in a circular path, what is the direction of the force that acts on the can
andreyandreev [35.5K]

Answer:

Toward the centre of the circular path

Explanation:

The can is moved in a circular path: this means that it is moving by circular motion (uniform circular motion if its tangential speed is constant).

In order to keep a circular motion, an object must have a force that pushes it towards the centre of the circular trajectory: this force is called centripetal force, and its magnitude is given by

F=m\frac{v^2}{r}

where m is the mass of the object, v its tangential speed, r the radius of the trajectory. This force always points towards the centre of the circular path.

3 0
3 years ago
On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around
Mekhanik [1.2K]

Answer:

v\approx 16.956\,\frac{m}{s}

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

\Sigma F = f = m\cdot \frac{v^{2}}{R}

The maximum speed is:

v = \sqrt{\frac{f\cdot R}{m} }

v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }

v\approx 16.956\,\frac{m}{s}

7 0
3 years ago
A high school physics student is sitting in a seat reading this question. The magnitude of the force with which the seat is push
Dennis_Churaev [7]
The force pushing down  is the force of Gravity. On a chair it is in perfect balance with the force pushing up (the normal force)
 

in terms of magnitude 
FN = FG = mg

the forces are in opposite direction

hope this helps
4 0
3 years ago
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