Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
<em>Values are gotten from the table named: blackbody radiati</em>on functions
<u>a) Calculate the band emission fractions for the visible region</u>
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
Answer:
Z = 29.938Ω ∠22.04°
I = 2.494A
Explanation:
Impedance Z is defined as the total opposition to the flow of current in an AC circuit. In an R-L-C AC circuit, Impedance is expressed as shown:
Z² = R²+(Xl-Xc)²
Z = √R²+(Xl-Xc)²
R is the resistance = 4Ω
Xl is the inductive reactance = ωL
Xc is the capacitive reactance =
1/ωc
Given C = 12 μF, L = 6 mH and ω = 2000 rad/sec
Xl = 2000×6×10^-3
Xl = 12Ω
Xc = 1/2000×12×10^-6
Xc = 1/24000×10^-6
Xc = 1/0.024
Xc = 41.67Ω
Z = √4²+(12-41.67)²
Z = √16+880.31
Z = √896.31
Z = 29.938Ω (to 3dp)
θ = tan^-1(Xl-Xc)/R
θ = tan^-1(12-41.67)/12
θ = tan^-1(-29.67)/12
θ = tan^-1 -2.47
θ = -67.96°
θ = 90-67.96
θ = 22.04° (to 2dp)
To determine the current, we will use the relationship
V = IZ
I =V/Z
Given V = 12V
I = 29.93/12
I = 2.494A (3dp)
Answer:
#include <iostream>
#include <iomanip>
using namespace std;
class pointType
{
public:
pointType()
{
x=0;
y=0;
}
pointType::pointType(double x,double y)
{
this->x = x;
this->y = y;
}
void pointType::setPoint(double x,double y)
{
this->x=x;
this->y=y;
}
void pointType::print()
{
cout<<"("<<x<<","<<y<<")\n";
}
double pointType::getX()
{return x;
}
double pointType::getY()
{return y;
}
private:
double x,y;
};
int main()
{
pointType p2;
double x,y;
cout<<"Enter an x Coordinate for point ";
cin>>x;
cout<<"Enter an y Coordinate for point ";
cin>>y;
p2.setPoint(x,y);
p2.print();
system("pause");
return 0;
}
Answer:
1072 acre foot
1331424000 kg
Explanation:
1 feet has 12 inches, so 2 in is 0.167 feet.
1 km^2 has 1 million m^2.
1 acre is 4074 m^2.
So, 1 km is 247 acres.
Then 26 km^2 is 6422 acres.
So, the volume of water is
6422 * 0.167 = 1072 acre-foot
Since one cubic meter of water has 1000 kg
One inch is 25.4 mm = 0.0254 m
One feet is 12 * 0.0254 = 0.3048 m
An acre-feet has a volume of
4074*0.3048 = 1242 m^3
And that is a mass of water of
1242 * 1000 = 1242000 kg/acre-feet
Therefore the mass of rainwater in the town is of
1072 * 1242000 = 1331424000 kg = 1331424 tons