Answer:
<em>d. unchanged.</em>
Explanation:
The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.
In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from
v = fλ
that the frequency is tied to the wave, and does not change throughout the waveform.
where v is the speed of the sound wave
f is the frequency
λ is the wavelength of the sound wave.
Answer:
The width of the strand of hair is 1.96 10⁻⁵ m
Explanation:
For this diffraction problem they tell us that it is equivalent to the diffraction of a single slit, which is explained by the equation
<h3> a sin θ =± m λ
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Where the different temrs are: “a” the width of the hair, λ the wavelength, θ the angle from the center, m the order of diffraction, which is the number of bright rings (constructive diffraction)
We can see that the diffraction angle is missing, but we can find it by trigonometry, where L is the distance of the strand of hair to the observation screen and "y" is the perpendicular distance to the first minimum of intensity
L = 1.25 m 100 cm/1m = 125 cm
y = 5.06 cm
Tan θ = y/L
Tan θ = 5.06/125
θ = tan⁻¹ ( 0.0405)
θ = 2.32º
With this data we can continue analyzing the problem, they indicate that they measure the distance to the first dark strip, thus m = 1
a = m λ / sin θ
a = 1 633 10⁻⁹ 1.25/sin 2.3
a = 1.96 10⁻⁵ m
a = 0.0196 mm
The width of the strand of hair is 1.96 10⁻⁵ m
Answer:
C) No work is required to move the negative charge from point A to point B.
Explanation:
An equipotential surface is defined as a surface connecting all the points at the same potential.
Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.
The work done when moving a charge is given by
where
q is the charge
is the potential difference between the initial and final point of motion of the charge
However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so
And so, the work done is also zero.
Answer:
The path of an object in uniform motion is a straight line.
Answer: D
Explanation:
Kinetic energy = 1/2mV^2
From the formula above, we can deduce that kinetic energy is proportional to the square of speed. That is,
K.E = V^2
Graphically, the relationship isn't linear but a positive exponential. Therefore, option D is the correct answer.
