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r-ruslan [8.4K]

2 years ago

11

Two strings are respectively 1.00 m and 2.00 m long. Which of the following wavelengths, in meters, could represent harmonics pr

esent on both strings?
1) 0.800, 0.670, 0.500
2) 1.33, 1.00, 0.500
3) 2.00, 1.00, 0.500
4) 2.00, 1.33, 1.00
5) 4.00, 2.00, 1.0

1 answer:

MrRissso [65]2 years ago

6 0

Answer:

5) 4.00, 2.00, 1.0

Explanation:

wave equation is given as;

F₀ = V / λ

Where;

F₀ is the fundamental frequency = first harmonic

Length of the string for first harmonic is given as;

L₀ = (¹/₂) λ

λ = 2 L₀

when L₀ = 1

λ = 2 x 1 = 2m

when L₀ = 2m

λ = 2 x 2 = 4m

For First harmonic, the wavelength is 2m, 4m

For second harmonic;

L₁ = (²/₂)λ

L₁ = λ

When L₁ = 1

λ = 1 m

when L₁ = 2

λ = 2 m

For second harmonic, the wavelength is 1m, 2m

Thus, the wavelength that could represent harmonics present on both strings is 4m, 2m, 1 m

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Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 × $10^{14}$ Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

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here E is the energy of electron per atom $E=\frac{x}{N}$ and h is plank constant i.e. $6.626\times10^{-34} Js$ and x is binding energy

and here N is the Avogadro constant = $6.023\times10^{23}$

so E will $\frac{x}{6.023\times10^{23}}$

E = $3.19\times10^{-19} J$

so put value in equation 1 we get

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1 year ago

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

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Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

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1 year ago

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Area of each wire, $A_1=A_2=A=\pi r^2=\pi (0.5\times 10^{-3})^2m^2$

Resistivity is the property of material due to which it offers resistance to the flow of current.

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1 year ago

Read 2 more answers

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Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

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Given that,

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Speed $v = v_{o}$

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We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

$\Delta W=\Delta K.E$

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If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

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1 year ago

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6 0

1 year ago