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:
1.693242
Explanation:
The colors in the Light emitting diodes have been identified by wavelength which is measured in nano-meters. Wavelength is a function of LED chip material. The LED diode which has a = 632 then A1 will be 1.63242, this is calculated by 1 / 632. Wavelength are important for human eye sensitivity. The colors emitted from the LED will depend on the semiconductor material.
Answer:
The term Accuracy means that how close our result to the original result.
Suppose we do any experiment in laboratory and we calculate mass = 7 kg but answer is mass = 15 kg then our answer is not accurate.
And the term Precision means how likely we get result like this.
Suppose we do any experiment in laboratory and we calculate mass five times and each time we get mass = 7 kg then our answer is precised but not accurate.
Answer:
Temperature on the inside ofthe box
Explanation:
The power of the light bulb is the rate of heat conduction of the bulb,
The thickness of the wall, L = 1.2 cm = 0.012m
Length of the cube's side, x = 20cm = 0.2 m
The area of the cubical box, A = 6x²
A = 6 * 0.2² = 6 * 0.04
A = 0.24 m²
Temperature of the surrounding,
Temperature of the inside of the box,
Coefficient of thermal conductivity, k = 0.8 W/m-K
The formula for the rate of heat conduction is given by:
Answer:
a) 4 passes are required to sort the string.
b) 4
c) i) TARP
ii) CHIP
iii) PART
iv) TARP
v) TARP
d) O(k+n), n is no. of strings, k is largest no. of character in among the string
O(d*(n+10)), n is no. of integers
Explanation: