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:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
 
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
 
        
             
        
        
        
Answer:
A&C
Explanation:
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talking with a friend can helping
 
        
                    
             
        
        
        
Answer:if power factor =1 is possible for that.
Explanation:when pf is unity. means 1.