Plis 3 conclusiones de este video

Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t

densk [106]

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t) be a real value signal for x ( jw ) = 0 , |w| > 200 $\pi$

g(t) = x ( t ) sin ( 2000 $\pi t )$

$x_{1} (t) = \frac{1}{2} x(t) sin ( 4000\pi t )$

next we apply Fourier transform

attached below is the remaining part of the solution

6 0

3 years ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

evablogger [386]

Answer:

Compute the number of gold atoms per cubic centimeter = 9.052 x 10^21 atoms/cm3

Explanation:

The step by step and appropriate substitution is as shown in the attachment.

From number of moles = Concentration x volume

number of moles = number of particles/ Avogadro's number

Volume = mass/density, the appropriate derivation to get the number of moles of atoms

5 0

3 years ago

What does the supply chain management process involve

VMariaS [17]

Answer:

It involves the active streamlining of a business's supply-side activities to maximize customer value and gain a competitive advantage in the marketplace

Explanation:

Supply chain management is the management of the flow of goods and services and includes all processes that transform raw materials into final products.

5 0

3 years ago

-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63

Fudgin [204]

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0

2 years ago

1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-

Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * $\frac{79.25}{21.2}$

= 0.0214 * 3.738

= 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0

3 years ago