Ask question

Ask question

All categories

2 years ago

12

Show that a chirped Gaussian pulse is compressed initially inside a single-mode fiber when ftC < 0. Derive expressions for th

e minimum width and the fiber length at which the minimum occurs.

1 answer:

Damm [24]2 years ago

8 0

Answer:

Given in diagram

You might be interested in

) A given system has four sensors that can produce an output of 0 or 1. The system operates properly when exactly one of the sen

Rashid [163]

A given system has four sensors that can produce an output of 0 or 1. The system operates proper . An alarm must be raised when two or more sensors have the output of 1. Design the simplest circuit that can be used to raise the alarm ly when exactly one of the sensors has its output equal to Repeat problem #4 for a system that has 7 sensors. Hint: Before you slog through a truth table with 128 rows in it, think about whether SOP or POS might be a better approach.

7 0

2 years ago

Why is it so dangerous to use a ground lift on a metal cased power tool

MaRussiya [10]

Answer:

Removal of the safety ground connection on equipment can expose users to an increased danger of electric shock and may contradict wiring regulations. ... If a fault develops in any line-operated equipment, cable shields and equipment enclosures may become energized, creating an electric shock hazard.

7 0

1 year ago

The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

4 0

2 years ago

As the porosity of a refractory ceramic brick increases:

ivolga24 [154]

Answer:

A) strength decreases, chemical resistance decreases, and thermal insulation increases

Explanation:

Strength always decreases, chemical resistence decreases, and thermal condictivity must be reduced therefore themal insulation must increase.

7 0

2 years ago

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

bagirrra123 [75]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.

Explanation:

Given:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

To calculate:

The magnitude of applied stress in the direction of [101] and [011].

Formula:

zcr=σ cosФ cosλ

Solution:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.

6 0

2 years ago