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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:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

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c) 53.13°

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S = v_{r} t

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c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

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cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

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Hello,

The answer is option A "Venus".

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pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
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<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

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<span>0 = (Vo×sin(α) + g×t/2)×t </span>
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<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
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<span>t = time = -2×Vo×sin(α)/g </span>
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<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
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<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
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<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
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Tanzania [10]
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