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alexandr1967 [171]
2 years ago
12

You have a string with a mass of 0.0135 kg. You stretch the string with a force of 8.29 N, giving it a length of 1.83 m. Then yo

u vibrate the string transversely at precisely the frequency that corresponds to its fourth normal mode, that is, at its fourth harmonic. What is the wavelength of the standing wave you create in the string
Physics
1 answer:
Sphinxa [80]2 years ago
7 0

The wavelength of the standing wave at fourth harmonic is; λ = 0.985 m and the frequency of the wave at the calculated wavelength is; f = 36.84 Hz

Given Conditions:

mass of string; m = 0.0133 kg

Force on the string; F = 8.89 N

Length of string; L = 1.97 m

1. To find the wavelength at the fourth normal node.

At the fourth harmonic, there will be 2 nodes.

Thus, the wavelength will be;

λ = L/2

λ = 1.97/2

λ = 0.985 m

2. To find the velocity of the wave from the formula;

v = √(F/(m/L)

Plugging in the relevant values gives;

v = √(8.89/(0.0133/1.97)

v = 36.2876 m/s

Now, formula for frequency here is;

f = v/λ

f = 36.2876/0.985

f = 36.84 Hz

Read more about Harmonics of standing waves at; brainly.com/question/10274257

#SPJ4

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Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

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Diameter of inlet pipe = 5.2 cm

Fifteen percent of the flow leaves through location (2)  and the remainder leaves at (3)

The mass flow rate is

m_{2}=0.15\times2

We need to calculate the mass flow rate at reach exit

Using formula of mass

m_{3}=m_{1}-m_{2}

m_{3}=2-0.15\times2

m_{3}=1.7\ kg/s

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v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}

v=21.9\ m/s

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The mass flow rate at reach exit is 1.7 kg/s.

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A pendulum is observed to complete 20 full cycles in 60 seconds. Determine the period and the frequency of the pendulum.
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i hope i have been useful buddy.

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Carbon and oxygen

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A light ray moving from oil (n=1.38) into an unknown material. If the light is incident at 35 and refracts at 23, what is the in
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The index of refraction of the unknown material in which a ray of light is incident at 35° and refracted at 23° is 2.03

<h3>Snell's law</h3>

index of refraction (n) = Sine i / Sine r

n = Sine i / Sine r

Where

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  • r is the angle of refraction
<h3>How to determine the refractive index </h3>

From the question given above, the following data were obtained:

  • Index of refraction of oil (nₒ) = 1.38
  • Angle of incidence (i) = 35°
  • Angle of refraction (r) = 23°
  • Index of refraction of unknown material (nᵣ) =?

nₒSine i = nᵣSine r

1.38 × Sine 35 = nᵣ × Sine 23

Divide both side by Sine 23

nᵣ = (1.38 × Sine 35) / Sine 23

nᵣ = 2.03

Thus, the index of refraction of the unknown material is 2.03

Learn more about Snell's law:

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