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valentina_108 [34]

3 years ago

15

Determine an expression in standard form for the voltage gain VoVs. Hv(jω)=Vo(jω)Vi(jω)=R2R111+jωCR2 Hv(jω)=Vo(jω)Vi(jω)=−R2R111

+jωCR2 Hv(jω)=Vo(jω)Vi(jω)=−R2R111−jωCR2 Hv(jω)=Vo(jω)Vi(jω)=−R1R211+jωCR2 HintsReferenceseBook & Resources Hint #1

1 answer:

IceJOKER [234]3 years ago

5 0

Answer:

Applying Kirchhoff's current law at inverting terminal

Explanation:

Detailed explanation is given in the attached document.

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For a body moving with simple harmonic motion state the equations to represent: i) Velocity ii) Acceleration iii) Periodic Time

max2010maxim [7]

Answer with Explanation:

The general equation of simple harmonic motion is

$x(t)=Asin(\omega t+\phi)$

where,

A is the amplitude of motion

$\omega$ is the angular frequency of the motion

$\phi$ is known as initial phase

part 1)

Now by definition of velocity we have

$v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )$

part 2)

Now by definition of acceleration we have

$a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )$

part 3)

The angular frequency is related to Time period 'T' as $T =\frac{2\pi }{\omega }$

where

$\omega$ is the angular frequency of the motion of the particle.

Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is $A\omega ^{2}$ and $A\omega$ respectively.

4 0

3 years ago

An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°

Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

$D = 7.5 \times 10^{-11} m^2/s$

for 0.225% carbon concentration following formula is used

$\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})$

where, erf stand for error function

$\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248$

$0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})$

$erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248$

$erf(\frac{x}{2\sqrt{DT}}) = 0.751$

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

$\frac{x}{2\sqrt{DT}} = 0.815$

$x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)$

$x = 2.39\times 10^{-3} m$

x = 0.002395 mm

8 0

3 years ago

1. Calculate the battery life in years when a pacemaker has the following characteristics: Battery Ampere-hours = 1.5 Pulse volt

Wittaler [7]

Answer:

battery life in year = 9 years and 48 days

Explanation:

given data

Battery Ampere-hours = 1.5

Pulse voltage = 2 V

Pulse width = 1.5 m sec

Pulse time period = 1 sec

Electrode heart resistance = 150 Ω

Current drain on the battery = 1.25 µA

to find out

battery life in years

solution

we get first here duty cycle that is express as

duty cycle = $\frac{width}{period}$ ...............1

duty cycle = 1.5 × $10^{-3}$

and applied voltage will be

applied voltage = duty energy × voltage ...........2

applied voltage = 1.5 × $10^{-3}$ × 2

applied voltage = 3 mV

so current will be

current = $\frac{applied\ voltage}{resistance}$ ................3

current = $\frac{3}{150}$

current = 20 µA

so net current will be

net current = 20 - 1.25

net current = 18.75 µA

so battery life will be

battery life = $\frac{1.5}{18.75*10^{-6}}$

battery life = 80000 hours

battery life in year = $\frac{80000}{8760}$

battery life in year = 9.13 years

battery life in year = 9 years and 48 days

4 0

3 years ago

A motor car shaft consists of a steel tube 30 mm internal diameter and 4 mm thick. The engine develops 10 kW at 2000 r.p.m. Find

tresset_1 [31]

The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.

<h3>What is power?</h3>

Power is the energy transferred per unit time.

Torque is find out by

P = 2πNT/60

10000 = 2π x 2000 x T / 60

T =47.74 N.m

The gear ratio Ne / Ns =4/1

Ns =2000/4 = 500

Ts =Ps x 60/(2π x 500)

Ts =190.96 N.m

Maximum shear stress τ = 16/π x (T / (d₀⁴ - d₁⁴))

τ max =T/J x D/2
where d₁ = 30mm = 0.03 m

d₀ = 30 +(2x 4) = 38mm =0.038 m

Substitute the values into the equation, we get

τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)

τ max = 28.98 MPa.

Thus, the maximum shear stress in the tube is 28.98 MPa.

Learn more about power.

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7 0

2 years ago

The given family of functions is the general solution of the differential equation on the indicated interval.Find a member of th

Alja [10]

Answer:

Explanation:

$y'''+y=0---(i)$

General solution

$y=c_1e^o^x+c_2\cos x +c_3 \sin x\\\\\Rightarrow y=c_1+c_2 \cos x+c_3 \sin x---(ii)\\\\y(\pi)=0\\\\\Rightarrow 0=c_1+c_2\cos (\pi)+c_3\sin (\pi)\\\\\Rightarrow c_1-c_2=0\\\\c_1=c_2---(iii)$

$y'=-c_2\cos x+c_3\cosx\\\\y'(\pi)=2\\\\\Rightarrow2=-c_2\sin(\pi)+c_3\cos(\pi)\\\\\Rightarrow-c_2(0)+c_3(-1)=2\\\\\Rightarrow c_3=-2\\\\y''-c_2\cos x -c_3\sin x\\\\y''(\pi)=-1\\\\\Rightarrow-1=-c_2 \cos (\pi)=c_3\sin(\pi)\\\\\Rightarrow-1=c_2-0\\\\\Rightarrow c_2=-1$

in equation (iii)

$c_1=c_2=-1$

Therefore,

$\large\boxed{y=-1-\cos x-2\sin x}$

5 0

3 years ago