boiiii I wont show you nun
Answer:
20 m/s
Explanation:
The speed of a wave is given by:
where
is the wavelength
f is the frequency
v is the speed
For the wave in this problem,
f = 10 Hz is the frequency
is the wavelength
So the speed is
The answer is λ₂ = 6.48 cm or 6.52 cm.
The out-of-tune guitar may have a wavelength between "6.48 cm" and "6.52 cm."
fb = |f2 − f1|
f₁ = 343/0.064
= 5276Hz
f₂ = 5276.9 Hz ± 17 Hz
f₂ = 5293.9 Hz or 5259.9 Hz
Now, calculating the possible wavelengths:
λ = 343/ 5259.9 or 343/ 5293.9
λ₂ = 6.48 cm or 6.52 cm
<h3>Why is beat frequency important?</h3>
When two waves with almost identical frequencies traveling in the same direction collide at a certain location, beats are produced. The opposing beneficial and harmful disruption causes the sound to alternatively be loud and weak whenever two sound waves with different frequencies reach your ear. This is referred to as beating.
The entire value of the frequency difference between the two waves is the beat frequency.
The following formula yields the beat frequency:
fb = |f2 − f1|
Learn more about beat frequency here:
brainly.com/question/14705053
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Answer:
That isnt a question so no one will know the answer to what you are talking about. I suggest adding a sceenshot or picture of the question.
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
