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3 years ago

For the given network, if R = 0.8 [Ω], the natural response of i in(t) is:

Engineering

1 answer:

11111nata11111 [884]3 years ago

6 0

Answer:

Iin(t) =1.3 × (-0.8/1.56) e^-t/1.56 A

Explanation:

In physics, the determination of the term " natural response" simply means that we want to know what happens in a circuit when the value of t = 0, that is to say after the circuit has been disconnected. Hence, the value of the voltage and the current can then be determined or Calculated;

For the the natural response of i in(t) we will be using the formula below;

I(t) = Vo × t/ R = Vo/R × e^-t/h.

Where h = 1/RC = time constant.

For t= 0^- = 0.8 × 1= 0.8 V.

1/Ctotal = 1/ 2 + 1/3 = 6/5

For t = 0^+;

h =( 0.8 + 0.5) × 6/5 = 1.56 seconds.

Hence, we will have;

Vin(t) = 0.8 × e^-t/1.56.

Iin(t) =1.3 × (-0.8/1.56) e^-t/1.56

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Why might a hospital patient prefer to interact with a person instead of robot?

MrMuchimi

A person would prefer to be able for the doctor/ nurse to visually see their symptoms and the problem, which a robot can not do, which would lead to a misdiagnosis

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3 years ago

assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2

amm1812

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

4 0

3 years ago

A wooden pallet carrying 540kg rests on a wooden floor. (a) a forklift driver decides to push it without lifting it.what force m

kicyunya [14]

Answer:

The appropriate solution is "1481.76 N".

Explanation:

According to the question,

Mass,

m = 540 kg

Coefficient of static friction,

$\mu_s$ = 0.28

Now,

The applied force will be:

⇒ $F=\mu_s mg$

By substituting the values, we get

$=0.28\times 540\times 9.8$

$=1481.76 \ N$

8 0

2 years ago

Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3

Neko [114]

Answer:

$\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

Explanation:

Diffusion flux of a gas, J is given by

$J=P_m\frac {\triangle P}{\triangle x}$ where $P_m$ is permeability coefficient, $\triangle$ P is pressure difference and x is thickness of membrane.

The pressure difference will be 10,000 Pa- 3000 Pa= 7000 Pa

At 298 K, the permeability coefficient of water vapour through polypropylene sheet is $38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)$

Since the thickness of sheet is given as 1mm= 0.1 cm then

$J=38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)\times \frac {7000 pa}{0.1cm}=0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}$

Therefore, the diffusion flux is $\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

7 0

3 years ago

Reference Parameters (returning multiple values): Write a C++ function that converts standard time to military time. Inputs incl

valkas [14]

Answer:

Code is given as below:

Explanation:

#include <iostream>

using namespace std;

//function prototype declaration

void MilitaryTime(int, int, char, int &, int &);

int main()

{

//declare required variables

int SHour, SMin, MHour, MMin;

char AorP;

//promt and read the hours from the user

cout<<"Enter hours in standard time : ";

cin>>SHour;

//check the hours are valid are not

while(SHour<0 || SHour>12)

{

cout<<"Invalid hours for standard time. "

<<"Try again..."<<endl;

cout<<"Enter hours in standard time : ";

cin>>SHour;

}

//promt and read the minutes from the user

cout<<"Enter minutes in standard time : ";

cin>>SMin;

//check the minutes are valid are not

while(SMin<0 || SMin>59)

{

cout<<"Invalid minutes for standard time. "

<<"Try again..."<<endl;

cout<<"Enter minutes in standard time : ";

cin>>SMin;

}

//promt and read the am or pm from the user

cout<<"Enter standard time meridiem (a for AM p for PM): ";

cin>>AorP;

//check the meridiem is valid are not

while(!(AorP=='a' || AorP=='p' || AorP=='A' || AorP=='P'))

{

cout<<"Invalid meridiem for standard time. "

<<"Try again..."<<endl;

cout<<"Enter standard time meridiem (a for AM p for PM): ";

cin>>AorP;

}

//call function to calculate the military time

MilitaryTime(SHour, SMin, AorP, MHour, MMin);

//fill zeros and display standard time

cout.width(2);

cout.fill('0');

cout<<SHour<<":";

cout.width(2);

cout.fill('0');

cout<<SMin;

if(AorP=='a' || AorP=='A')

cout<<" am = ";

else

cout<<" pm = ";

//fill zeros and display military time

cout.width(2);

cout.fill('0');

cout<<MHour;

cout.width(2);

cout.fill('0');

cout<<MMin<<endl;

system("PAUSE");

return 0;

}

//function to calculate the military time with reference parameters

void MilitaryTime(int SHour, int SMin, char AorP, int &MHour, int &MMin)

{

//check the meredium is am or pm

//and calculate hours

if(AorP=='a' || AorP=='A')

{

if(SHour==12)

MHour = 0;

else

MHour = SHour;

}

else

MHour = SHour+12;

MMin = SMin;

5 0

3 years ago