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irakobra [83]
3 years ago
14

An amplifier with 40 dB of small-signal, open-circuit voltage gain, an input resistance of 1 MO, and an output resistance of 100

Q. drives a load of 500 Q. What voltage and power gains (expressed in dB) would you expect with the load connected
Engineering
1 answer:
Vanyuwa [196]3 years ago
3 0

convert 40db to standard gain

AL=10^40/20=100

calculate total voltage gain

=AL×RL/RL+Ri

=83.33

38.41 DB

calculate power

Pi=Vi^2/Ri Po=Vo^2/RL

power gain= Po/Pi

=13.90×10^6

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Why or why not the following materials will make good candidates for the construction of
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Answer:

Answer explained below

Explanation:

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3 0
3 years ago
Which statement most accurately describes Pascal's law?
marin [14]

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La  C

Explanation:

7 0
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A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
When the psychologist simply records the relationship between two variables...
Wewaii [24]
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8 0
3 years ago
A battery is connected to a resistor. Increasing the resistance of the resistor will __________. A battery is connected to a res
belka [17]

Answer: the increase in the external resistor will affect and decrease the current in the circuit.

Explanation: A battery has it own internal resistance, r, and given an external resistor of resistance, R, the equation of typical of Ohm's law giving the flow of current is

E = IR + Ir = I(R + r)........(1)

Where IR is the potential difference flowing in the external circuit and Or is the lost voltage due to internal resistance of battery. From (1)

I = E/(R + r)

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8 0
3 years ago
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