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fiasKO [112]
3 years ago
8

Now, suppose that you have a balanced stereo signal in which the left and right channels have the same voltage amplitude, 500 mV

pp. This time, however, you want to be able to mix these two channels into a single inverted output while independently varying the gain of the two channels. Design and build an op amp circuit with potentiometers so that you can independently vary the gain of the left and right channels. Choose resistors so that the overall output of your circuit ranges between 0.1Vpp (when both channels are set to minimum gain) and 20Vpp (when both channels are set to maximum gain).

Engineering
1 answer:
eimsori [14]3 years ago
6 0

Answer:

R₁ = 32kΩ

Explanation:

See attached image

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Fittoniya [83]

Answer:

LibreOffice is an integrated suite of software applications used to perform office tasks such as word processing, presentation preparation and spreadsheet calculations

4 0
3 years ago
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A scrubber on a coal-fired power plant is useful to remove
Eddi Din [679]

Answer:

sulfur dioxide

Explanation:

The scrubber is an apparatus installed in a coal-fired power plant to clean the passing gas through the smokestack.  Due to the norm enacted through the clean air Act, almost all the scrubber used in the U.S is used to remove sulfur concentration from coal. it can remove approx 90-95% SO_2 from the smokestack.

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3 years ago
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Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is
VARVARA [1.3K]

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}

Solve the following for the real stress and pressure for the stable.\sigma_{r1}=K(\varepsilon_{r1})^{n}

K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}

Solve the following for the true state stress and stress2.

\sigma_{r2}=K(\varepsilon_{r2})^n

     =\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa

4 0
3 years ago
Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t
alukav5142 [94]

Answer:

  • hoop stress
  • longitudinal stress
  • material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = \frac{PD}{2T}     EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = \frac{PD}{4T}       EQUATION 2

where p = water pressure inside the hose

          d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

\frac{D}{T} ≥  20

insert this value into equation 1

αh = \frac{20 *30}{2}  = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

8 0
3 years ago
In a reversing 2-high mill, a series of cold rolling process is used to reduce the thickness of a plate from 45mm down to 20mm.
podryga [215]

Answer:

Explanation:

Given

Initial Thickness=45 mm

Final thickness=20 mm

Roll diameter=600 mm

Radius(R)=300 mm

coefficient of friction between rolls and strip (\nu)=0.15

maximum draft(d_{max})=\nu ^2R

=0.15^2\times 300=6.75 mm

Minimum no of passes=\frac{45-20}{6.75}=3.70\approx 4

(b)draft per each pass

d=\frac{Initial\ Thickness-Final\ Thickness}{min.\ no.\ of\ passes}

d=\frac{45-20}{4}=6.25 mm

5 0
3 years ago
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