Answer:
Detailed solution is attached below in three simple steps the problem is solved.
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
Answer:
V = 6.33 m/s
Explanation:
Given:
- The length of the wire L = 0.02 m
- The diameter of the wire D = 0.0005 m
- The calibration expression V = 0.0000625*h^2
- Environment temperature T_inf = 298 K
- Surface temperature T_s = 348 K
- The voltage drop dV = 5 V
- The electric current I = 0.1 A
Find:
- the velocity of Air
Solution:
- Calculate the surface area of the wire:
A = pi*D*L
A = pi*(0.0005)*(0.02) = 0.00003142 m^2
- The rate of energy in the wire P:
P = I*dV = 0.1*5 = 0.5 W
- Apply Newton's Law of Cooling:
P = h*A*(T_s - T_inf)
h = P /A*(T_s - T_inf)
Plug in the values:
h= 0.5/ 0.00003142*(348 - 298)
h = 318.27 W /m^2K
- Using the calibration relationship given, compute the velocity of air:
V = 6.25*10^-5 * h^2
V = 6.25*10^-5 * (318.27)^2
V = 6.33 m/s
Answer:
#include<iostream>
using namespace std;
int main()
{
double data[3][5],avg,least,most,total;
int leastNum,mostNum;
for(int i=0;i<3;i++)
{
cout<<"Enter quantity of food for monkey "<<i+1<<":\n";
for(int j=0;j<5;j++)
{
cout<<"Day "<<j+1<<": ";
cin>>data[i][j];
}
}
least = data[0][0];
most = data[0][0];
for(int i = 0;i<3;i++)
{
for(int j = 0;j<5;j++)
{
total+=data[i][j];
if(data[i][j]>most)
{
most=data[i][j];
mostNum=i+1;
}
if(data[i][j]<least)
{
least=data[i][j];
leastNum=i+1;
}
}
}
avg=total/5.0;
cout<<"Average food eaten in a day by all the 3 monkeys: "<<avg<<endl;
cout<<"Most amount of food eaten in a day: "<<most<<" by monkey: "<<mostNum<<endl;
cout<<"Least amount of food eaten in a day: "<<least<<" by monkey: "<<leastNum<<endl;
return 0;
}
Answer:
yes you got it correct
Explanation:
hope you have a good day and stay positive ❤️
