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2 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]2 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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An old refrigerator consumes 247 W of power. Assuming that the refrigerator operates for 19 hours everyday, what is the annual o

german

Answer:

The annual operating cost of the refrigerator is $102.78.

Explanation:

Power consumed by the refrigerator = 247 W = 247/1000 = 0.247 kW

Daily operation of the refrigerator = 19 hours

Annual operation of the refrigerator = 365 × 19 = 6,935 hours

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1712.945 kWh will cost 1712.945 × $0.06 = $102.78

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The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to

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Answer:

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minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N

we will calculate the value of by applying this formula

= $\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12$ = 1.56 * 10^-5

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hence Z (section modulus ) = BD^2 / 6

= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8

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= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )

= 0.247 N/mm2

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3 years ago

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A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c

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Answer:

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$101.96\ \text{lb/ft}^2$

Explanation:

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$P_1=0$

$h_1=h_2$

From Bernoulli's law we have

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The maximum pressure on the girl's hand is $10.8\ \text{lb/ft^2}$

Now $v_1$ = 200 mph = $200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}$

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