Given:

frequency, f = 60.0 Hz
frequency, f' = 45.0 Hz

Solution:
To calculate max current in inductor,
:
At f = 60.0 Hz


L = 0.1326 H
Now, reactance
at f' = 45.0 Hz:


Now,
is given by:
Therefore, max current in the inductor,
= 2.13 A
Answer:
15,000 psi
Explanation:
The solution / solving is attach below.
Answer:
15.8
0.0944
Explanation:
L = 1.5
B = 1.0
Speed of water = 15cm
Temperature = 20⁰C
At 20⁰C
Specific weight = 9790
Kinematic viscosity v = 1.00x10^-4m²/s
Dynamic viscosity u = 1.00x10^-3
Density p = 998kg/m²
Reynolds number
= 0.15x1.5/1.00x10^-4
= 225000
S = 5
5x1.5/225000^1/2
= 0.0158
= 15.8mm
Resistance on one side of plate
F = 0.664x1x1.0x10^-3x0.15x225000^1/2
= 0.04724N
Total resistance
= 2N
= 2x0.04724
= 0.0944N
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