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katrin2010 [14]

3 years ago

12

An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side

frequencies. What is the total bandwidth of the AM signal

1 answer:

Aliun [14]3 years ago

6 0

Answer:

total width bandwidth = 8kHz

Explanation:

given data

transmitter operating = 3.9 MHz

frequencies up to = 4 kHz

solution

we get here upper side frequencies that is

upper side frequencies = 3.9 × $10^{6}$ + 4 × 10³

upper side frequencies = 3.904 MHz

and

now we get lower side frequencies that is

lower side frequencies = 3.9 × $10^{6}$ - 4 × 10³

lower side frequencies = 3.896 MHz

and now we get total width bandwidth

total width bandwidth = upper side frequencies - lower side frequencies

total width bandwidth = 8kHz

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zhuklara [117]

Answer:

I think option d is the answer

6 0

3 years ago

#198. Moment of inertia about center of a segmented bar A bar of width is formed of three uniform segments with lengths and area

zaharov [31]

Complete Complete

The complete question is shown on the first uploaded image

Answer:

The moment of inertia of the bar about the center of mass is

$I_r = 1888.80 \ kg m^2$

Explanation:

The free body diagram is shown on the second uploaded image

From the diagram we see that is

The mass of each segment is

$m_1 = \rho_1 l_1 w = 1 * 6 * 2 = 12$

$m_1 = \rho_2 l_2 w = 8 * 6 * 2 = 96$

$m_1 = \rho_2 l_2 w = 5 * 5 * 2 = 50$

The distance from the origin to the center of the segments i.e the center of masses for the individual segments

$x_2 = \frac{6}{2} + 6 = 9 m$

$x_3 = \frac{4}{2} + 12 = 14 m$

The resultant center of mass is mathematically evaluated as

$x_r = \frac{m_1 * x_1 + m_2 *x_2 + m_3 *x_3}{m_1 + m_2 + m_3}$

$= \frac{12 * 3 + 96 *9 + 50 *14}{12+ 96 + 50}$

$x_r = 10.13m$

The moment of Inertia of each segment of the bar is mathematically evaluated

$I_1 =\frac{m_1}{12}(l_1^2 + w^2)$ = $\frac{12}{12}(1^2 + 2^2)$

$I_1 = 4 \ kgm^2$

$I_2 =\frac{m_2}{12}(l_2^2 + w^2)$ = $\frac{96}{12}(6^2 + 2^2)$

$I_2 = 320 \ kgm^2$

$I_3 =\frac{m_3}{12}(l_3^2 + w^2)$ = $\frac{50}{12}(4^2 + 2^2)$

$I_2 = 83.334 \ kgm^2$

According to parallel axis theorem the moment of inertia about the center ( $x_r$ ) is mathematically evaluated as

$I_r = (I_1 + m_1 r_1^2) + (I_2 + m_2 r_2^2) +(I_3 + m_3 r_3^2)$

$I_r = (I_1 + m_1 |x_r - x_1|^2) + (I_2 + m_2 |x_r - x_2|^2) +(I_3 + m_3 |x_r - x_3|^2)$

$I_r = (4 + 12 |10.13 - 3|^2) + (320 + 96 |10.13 - 9|^2) +(83.334 + 50 |10.13 - 14|^2)$

$I_r = 1888.80 \ kg m^2$

6 0

3 years ago

A long rod of 60-mm diameter and thermophysical properties rho=8000 kg/m^3, c=500J/kgK, and k=50 W/mK is initally at a uniform t

Monica [59]

Answer:

Tc = 424.85 K

Explanation:

Given that,

$D = 60 mm = 0.06 m$

$\rho = 8000 kg/m^3$

$k = 50 w/m . kc = 500 j/kg.k$

$h_{\infty} = 1000 w/m^2t_{\infity} = 750 kt_w = 500 K$

$surface area = As = \pi dL \\\frac{As}{L} = \pi D = \pi \timeS 0.06$

HEAT FLOW Q is

$Q = h_{\infty} As (T_[\infty} - Tw) = 1000 \pi\times 0.06 (750-500)$

= 47123.88 w per unit length of rod

volumetric heat rate

$q = \frac{Q}{LAs}$

$= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}$

$q = 1.66\times 10^{7} w/m^3$

$Tc = \frac{- qR^2}{4K} + Tw$

$= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 50} + 500$

= 424.85 K

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2 years ago

The news media often report an earthquake's magnitude on the Richter scale. Which of the following items are characteristics of

vlabodo [156]

The correct answers are A) determined from the amplitude recorded by a single seismometer. C) represents the amount of energy released. D) uses a logarithmic scale.

The news media often report an earthquake's magnitude on the Richter scale. The following items are characteristics of a Richter earthquake magnitude: determined from the amplitude recorded by a single seismometer, represents the amount of energy released, and uses a logarithmic scale.

When measuring the scale of the magnitude of earthquakes, the Richter scale measures the scale od moderated size earthquakes. It was invented by two United States seismologists: Charles F. Richer and Beno Gutenberg. They developed the measure in California when they were researching from 1927 through 1936, in the Seismological Laboratory of the Carnegie Institution of Washington, in Pasadena, California.

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3 years ago

A wing generates a lift L when moving through sea-level air with a velocity U. How fast must the wing move through the air at an

vredina [299]

Answer:

V1 = 1.721 * V2

Explanation:

To start with, we assume that both lift forces are equal, such that

L2 = L1

1 is that of the level at 10000 m, and 2 is that of the level at sea level.

Next, we try and substitute the general formula for both forces such that

C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A

On further simplification, we have

ρ1.V1² = ρ2.V2², making V1 subject of formula, we have

V1 = √(ρ2/ρ1). V2²

Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have

V1 = √(1.225/0.4135) * V2

V1 = √2.9625 * V2

V1 = 1.721 * V2

4 0

2 years ago