All categories

3 years ago

An example of the split-off point in oil, gasoline, and kerosene production is that point where crude oil is

Engineering

1 answer:

eimsori [14]3 years ago

8 0

i believe the correct answer is c but i’m sorry if i’m not correct

You might be interested in

1. Làm thế nào để đảm bảo tính khả thi của văn bản hành chính ?

Dennis_Churaev [7]

Answer:

Explanation:

......................

7 0

2 years ago

What is the steady-state value of the output of a system with transfer function G(s)= 6/(12s+3), subject to a unit-step input?

fenix001 [56]

Answer:

At steady state output will be 2

Explanation:

We have given transfer function $G(S)=\frac{6}{12S+3}$

Input is unit step so $X(S)=\frac{1}{S}$

We know that $G(S)=\frac{Y(S)}{X(S)}$ , here $Y(S)$ , is output

So output $Y(S)=G(S)\times X(S)$

$Y(S)=\frac{1}{S}\times \frac{6}{12S+3}$

Taking 12 common from denominator

$Y(S)=\frac{1}{2S(S+\frac{1}{4})}$

Now using partial fraction

$\frac{1}{2S(S+\frac{1}{4})}=\frac{A}{2S}+\frac{B}{(S+\frac{1}{4})}$

$\frac{1}{2S(S+\frac{1}{4})}=\frac{A(S+\frac{1}{4}+2BS)}{2S(S+\frac{1}{4})}$

$AS+\frac{A}{4}+2BS=1$

On comparing coefficient A=4 and B = -2

Putting the values of A and B in Y(S)

$Y(S)=\frac{4}{2S}-\frac{2}{S+\frac{1}{4}}$

Now taking inverse la place

$y(t)=2-2e^{\frac{-t}{4}}$

Steady state means t tends to infinite

So output at steady state = $y(t)=2-2e^{\frac{-\infty}{4}}$

$y(t)=2-0=2$

3 0

2 years ago

Four eight-ohm speakers are connected in parallel to an audio power amplifier. The amplifier can supply a maximum driver output

Vera_Pavlovna [14]

Answer:

112.5 watts

Explanation:

The output voltage (V) = 15 volts

The equivalent resistance for the speakers connected in parallel ( $R_T$ ) is gotten by using the formula:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\\\But\ R_1=R_2=R_3=R_4=8\ ohm\\\\\frac{1}{R_T} =\frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} \\\\\frac{1}{R_T} =\frac{1+1+1+1}{8} \\\\\frac{1}{R_T} =\frac{4}{8} \\\\R_T=\frac{8}{4} \\\\R_T=2\ ohm$

The current flowing through the amplifier (I) is:

I = V / $R_T$ = 15 / 2 = 7.5 A

The audio power (P) when outputting this maximum voltage is given by:

P = V² / $R_T$ = I² $R_T$ . Therefore:

P = V² / $R_T$ = 15² / $R_T$ = 112.5 watts

6 0

2 years ago

As you push a toggle bolt into a wall, the

MakcuM [25]

Answer:

Explanation:
One the object is attached the toggle it's placed onto the bolt. The wings should be open in the direction of the bolt head. The toggle is than collapsed and the entire toggle bolt is inserted into the hole in the wall. Once the wings are fully through the opening they will spring open on the other side.
Hopt it's help thanks..

5 0

1 year ago

Consider a step pn junction made of GaAs at T = 300 K. At zero bias, only 20% of the total depletion region width is in the p-si

Nat2105 [25]

Answer:

0.31 μm

Explanation:

this question wants us to Determine the depletion region width, xn, in the n-side in unit of μm. using the information below.

density in the p-side = 5.68x10^16

density in the n-side = 1.42x10^16

$\sqrt{\frac{2*12.7*8.85E-10}{1.6E-14}(\frac{1}{5.68E16}+\frac{1}{1.42E16} )(1.2) }$

= √(1.42x10⁵)(1.76056335x10⁻¹⁷ + 7.042253521x10⁻¹⁷)(1.2)

= √150.74x10⁻¹¹

= 3.882x10⁻⁵

approximately 0.39μm

xn = 0.39 x 0.8

= 0.31μm

0.31 um is the depletion region width. thank you!

3 0

2 years ago