3 years ago

Rose is building a circuit that accepts sound input and amplifies it. Which component will perform the amplification function?

Engineering

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

FET Transistor

Explanation:

A properly biased Field Effect Transistor will take the audio input and amplify it to drive speakers

You might be interested in

Think for a moment about the potential problems with big data that the speaker mentioned:

Natalka [10]

Answer: 1. sadly yes, some people are treated unfairly for crimes people yet have commited. 2. no 3. yes

Explanation:

i did this last year

5 0

3 years ago

Read 2 more answers

Explain mathematically whether the divergence of the curl of a vector field is

svp [43]

Answer:

yes

Explanation:

6 0

2 years ago

Is an isothermal process necessarily internally reversible? Explain your answer with an example

torisob [31]

Answer:

please give me brainlist and follow

Explanation:

Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.

5 0

2 years ago

What standard feature is on all circular saws?

PtichkaEL [24]

What standard feature is on all circular saws?

4. All of the above

4 0

3 years ago

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin

Leni [432]

Answer:

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

$p.f_1=0.77$

$cos \theta_1 =0.77$

$\theta_1=cos^{-1}0.77$

$\theta_1=39.65 \textdegree$

And

$p.f_2=0.9$

$cos \theta_2 =0.9$

$\theta_2=cos^{-1}0.9$

$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

6 0

2 years ago