Answer:
Angle θ = 30.82°
Explanation:
From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ
where;
I_o is the intensity of the polarized wave before passing through the filter.
In this question,
I is 0.708 W/m²
While I_o is 0.960 W/m²
Thus, plugging in these values into the equation, we have;
0.708 W/m² = 0.960 W/m² •cos²θ
Thus, cos²θ = 0.708 W/m²/0.960 W/m²
cos²θ = 0.7375
Cos θ = √0.7375
Cos θ = 0.8588
θ = Cos^(-1)0.8588
θ = 30.82°
Explanation:
Given that,
Magnitude of vector A, |A| = 15
Magnitude of vector B, |B| = 25
We need to find the magnitude of this sum.
The maximum sum of the resultant vector,
The minimum sum of the resultant vector,
So, the magnitude of this sum either 45 or -10.
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
Answer:
0.5m
Explanation:
v=f×lamda
v is 300m/s, f is 600Hz, lamda is ?
lamda=v/f
lamda=300/600
lamda =3/6=1/2m
The resistance between A and B is 10 ohms.
