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ale4655 [162]
3 years ago
14

A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What wil

l be the wavelength if the tension is decreased to 600 N and the frequency is kept constant
Physics
1 answer:
mash [69]3 years ago
5 0

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu =  \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\

when the tension is decreased to 600 N, that is T₂ = 600 N

T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2  = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2  = \sqrt{0.11466} \\\\\lambda _2  =0.339 \ m\\\\\lambda _2  =33.9  \ cm

Therefore, the wavelength will be 33.9 cm

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Answer:

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<em></em>

Explanation:

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Answer:

(a) 4.27 x 10^-4 Telsa

(b) 3.28 x 10^-4 Telsa

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B = \frac{\mu_{0}Ni}{2\pi r_{inner}}

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B = 4.27 x 10^-4 Telsa

(b)

The outer radius is R = 18.1 + 5.49 = 23.59 cm = 0.236 m

The magnetic field due to the outer radius is

B = \frac{\mu_{0}Ni}{2\pi r_{outer}}

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Explanation:

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