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:
A) Cancer of the Lungs
B)Larynx and Urinary Tract, as well as nervous system and kidney damage
Explanation:
Explanation:
what are the options for this question?
#include
int main () {
printf("Program to calculate the square footage of the house.\n");
int total_rooms;
double length, width;
double total_square_footage = 0.0;
printf("Enter total number of rooms in the house:");
scanf("%d", &total_rooms);
for (int i = 0; i
printf("Enter the lenght and width of room %d: ", i+1);
scanf("%if %if", &lenght, &width);
total_square_footage += lenght*width;
}
printf("Total square footage of the house: %if\n", total_square_footage);
return 0;
}
Please mark it as brainliest answer:).
Answer:
Depth in the contracted section = 2.896m
Velocity in the contracted section = 2.072m/s
Explanation:
Please see that attachment for the solving.
Assumptions:
1. Negligible head losses
2. Horizontal channel bottom
