The point of contact the path difference is zero but one of the interfering ray is reflected so the effective path difference becomes λ/2 thus the condition of minimum intensity is created in the center.
363 m/s is the speed of sound through the air in the pipe.
Answer: Option B
<u>Explanation:</u>
The formula used to calculate the wavelength given as below,
--------> eq. 1
In power system, harmonics define by positive integers of the fundamental frequency. So the third order harmonic is a multiple of the third fundamental frequency. Each harmonic creates an additional node and an opposite node, as well as an additional half wave within the string.
If the number of waves in the circuit is known, the comparison between standing wavelength and circuit length can be calculated algebraically. The general expression for this given as,
For first harmonic, n =1
For second harmonic, n =2
For third harmonic, n =3
-------> eq. 2
Here given f = 939 Hz, L = 0.58 m...And, substitute eq 2 in eq 1 and values, we get
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D)
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The average kinetic energy of the molecules that make up the solution also increases with increasing temperature. This increase in kinetic energy allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions. </span></span></span></span>
Answer:
the smallest angle from the antennas is <em>47.3°</em>
Explanation:
We first need to write the expression for the relation between the wavelength (λ) and the frequency (f) of the wave, and then solve for the wavelength.
Therefore, the relation is:
λ = c /f
where
- c is the speed of light constant
- λ is the wavelength
- f is the frequency
Thus,
λ = (3 × 10⁸ m/s) / (3.4 MHz)
= (3 × 10⁸ m/s) / (3.4 MHz)(10⁶ Hz/1 MHz)
= 88.235 m
Therefore, the smallest angle measured (from the north of east) from the antennas for the constructive interference of the two-radio wave can be calculated as
θ = sin⁻¹(λ / d)
where
- d is the distance between the two radio antennas
Thus,
θ = sin⁻¹(88.235 / 120)
<em>θ = 47.3 °</em>
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Therefore, the smallest angle from the antennas, measured north of east, at which constructive interference of two radio waves occurs is <em>47.3 °</em>.
