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3 years ago

PLS :(((( HELP HELPPPP

Engineering

1 answer:

borishaifa [10]3 years ago

7 0

Answer:

the awnser is A

Explanation:

the awnser is A

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4.71 A full-wave rectifier circuit with a 1-kΩ operates from a 120-V (rms) 60-Hz household supply through a 6-to-1 transformer h

brilliants [131]

Answer:

$V_{p (load)} = 28,3 V - 0,7 V = 27,6 V$

$V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V$

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

$I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA$

Explanation:

The peak voltage after the 6 to 1 step down is $V_{p} = \frac{120}{6} \sqrt{2} = 28,3V$ . Then, the peak voltage of the rectified output is $V_{p} - [tex]V_{avg} = \frac{2V_{p (load)} }{\pi} = \frac{55,2 V}{\pi } = 17,6 V$ V_{d}[/tex] and according to the statement, the diodes can be modeled to be $V_{d} = 0,7 V$ . Then, the peak voltage in the load is $V_{p (load)} = 28,3 V - 0,7 V = 27,6 V$ .

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

The average output voltage is calculated as:

$V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V$

The average current in the load is calculated as:

$I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA$

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3 years ago

prove that the heat transfer at the constant pressure is given by the enthalpy change during the process

mihalych1998 [28]

Answer:

at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH=qp.

Explanation:

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3 years ago

Which type of blade is used with a demolition saw?

Scrat [10]

The correct answer is An abrasive wheel.

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3 years ago

10. Which of these requires a wheel alignment after replacement?

Elden [556K]

C. Both; require a wheel alignment after replacement

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2 years ago

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A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude i

Gre4nikov [31]

Answer:

13.4 mm

Explanation:

Given data :

Load amplitude ( F ) = 22,000 N

factor of safety ( N )= 2.0

Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa

<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>

minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm

<em>attached below is a detailed solution</em>

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2 years ago