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3 years ago

5

An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM sig

nal. If it is desired to operate at 50% modulation, what must be the dc input power to the final intelligence signal amplifier

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

"192.3 watt" is the right answer.

Explanation:

Given:

Efficient amplifier,

= 65%

or,

= 0.65

Power,

$P_c=250 \ watt$

As we know,

⇒ $P_t=P_c(1+\frac{\mu^2}{2} )$

By putting the values, we get

$=P_c(1+\frac{1}{2} )$

$=1.5 \ P_c$

Now,

⇒ $P_i=(P_t-P_c)$

$=1.5 \ P_c-P_c$

$=\frac{P_c}{2}$

DC input (0.65) will be equal to " $(\frac{P_c}{2} )$ ".

hence,

The DC input power will be:

= $\frac{250}{2}\times \frac{1}{0.65}$

= $\frac{125}{0.65}$

= $192.3 \ watt$

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Based on the calculations, the absolute pressure in the pipe is equal to 90,670.4‬ N/m².

<h3>How to calculate the absolute pressure?</h3>

Mathematically, absolute pressure can be calculated by using this formula:

$P_{abs} = P_{atm} + P_{guage}$

<u>Scientific data:</u>

Atmospheric pressure (Patm) = 1 bar = 1 × 10⁵ N/m²

Head (h) = -70mmhg = -0.07 m hg

Acceleration due to gravity = 9.8 m/s²

Density of mercury = 13,600 kg/m³.

Next, we would determine the gauge pressure:

$P_{guage} = \rho gh\\\\P_{guage} = -13600 \times 9.8 \times 0.07\\\\P_{guage} = -9,329.6\;N/m^2$

Now, we can calculate the absolute pressure:

$P_{abs} = P_{atm} + P_{guage}\\\\P_{abs} = 1 \times 10^5 - 9329.6$

Absolute pressure = ‭90,670.4‬ N/m².

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2 years ago

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3 years ago

Suppose that the weights for newborn kittens are normally distributed with a mean of 125 grams and a standard deviation of 15 gr

kherson [118]

(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

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In a normal distribution curve;

2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%

1 standard deviation (d) below the mean (M), (M - d) is at 16 %
1 standard deviation (d) above the mean (M), (M + d) is at 84%
2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%

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95 g is at 2% and 110 g is at 16%

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Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g

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TEA [102]

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4 years ago

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