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2 years ago

(a) Design a first-order passive high-pass filter with a cutoff frequency of 1000 rad/sec.

Engineering

1 answer:

balu736 [363]2 years ago

6 0

Answer:

See the attachment

Verification shown in explanation

Explanation:

wc= 1/(RC)

The value of R and C shown in the attachment can be calculated as follows:

R=100 ohms

1000= 1/(100×C)

C= 10 microfarads

Gain= R/√(R²+Xc²)

at f=500, wc= 3142.59 rad/s

Xc= 1/(2πfC)

Xc=31.83

Gain= 100/√(100²+31.83²)

Gain= 0.953

at fc= 100, wc= 628.32 rad/s

Xc= 159.15

gain= 100/√(100²+ 159.15²)

gain= 0.532

As see form calculations that gain at low frequencies is smaller and gain at very high frequencies is near to 1. This means that at high frequencies Vin≈ Vout

Hence this filter allows high frequencies to pass but blocks low frequencies by lowering the gai

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An array of electronic chips is mounted within a sealedrectangular enclosure, and colling is implemented by attaching analuminum

Licemer1 [7]

Answer:

Base temperature is 46.23 °C

Explanation:

I've attached explanations

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2 years ago

A 10 hp motor is used to raise a 1000 Newton weight at a vertical distance of 5 meters. What is the work the motor performs?

Aleksandr [31]

The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.

Define work. Explain the rate of doing work.

Work is the energy that is moved to or from an item by applying force along a displacement in physics. For a constant force acting in the same direction as the motion, work is easiest expressed as the product of force magnitude and distance traveled.

Since the force transfers one unit of energy for every unit of work it performs, the rate at which work is done and energy is used are equal.

Solution Explained:

Given,

Weight = 1000N and distance = 5m

A/Q, the work here is done in lifting then

Work = (weight) × (distance moved)

= 1000 X 5

= 5000Nm or 5000J = 5kJ

Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.

To learn more about work, use the link given
brainly.com/question/25573309
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4 0

1 year ago

For powder compaction using a single-action punch, derive an expression for the distribution of axial pressure within a die of r

Zielflug [23.3K]

Answer:

a) 2∪p/lb (l+b)dH

b) po exp( 4∪x/l)

Explanation:

please check the attachment for proper explanation and proper sign notations thanks.

3 0

3 years ago

If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less

liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

$R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19} } =0.93ohm*cm$

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })$

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })$

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

$V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2} } , if n_{i}=1.5x10^{10}cm^{-3} \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15} }{(1.5x10^{10})^{2} } )=0.83V$