To solve this problem we will apply the concept related to destructive interference (from the principle of superposition). This concept is understood as a superposition of two or more waves of identical or similar frequency that, when interfering, create a new wave pattern of less intensity (amplitude) at a point called a node. Mathematically it can be described as
Where,
d = Path difference
= wavelength
n = Any integer which represent the number of repetition of the spectrum
In this question the distance between the two source will be minimum for the case of minimum path difference, then n= 1
Therefore the minimum distance that should you separate two sources emitting the same waves is 2.5mm
Answer:
80 amperes
Explanation:
Current in the circuit = ?
Voltage in the circuit = 160 Volts
Resistance = 2 Ω
Voltage = Current x Resistance
V = IR
160V = I x 2 Ω
I = 160V / 2 Ω
I = 80 Amperes
Therefore the current in the circuit is 80 amperes :)
Answer:
Sound waves are pushed closer together, decreasing wavelength
and increasing frequency.
Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
