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makkiz [27]
3 years ago
6

The mass of a string is 2.40×10-3 kg, and it is stretched so the tension in it is 120 N. A transverse wave traveling on this str

ing has a frequency of 260 Hz and a wavelength of 0.60 m. What is the length of the string?
Physics
1 answer:
ankoles [38]3 years ago
3 0

Answer:

Length of the string, l = 0.486 meters

Explanation:

It is given that,

Mass of the string, m=2.4\times 10^{-3}\ kg

Tension in the string, T = 120 N

Frequency of transverse wave, f = 260 Hz

Wavelength of the wave, \lambda=0.6\ m

The speed of a transverse wave (v) is given by :

v=\sqrt{\dfrac{T}{\mu}}........(1)

<em>Where,</em>

\mu=\dfrac{m}{l}

Also, speed of a wave, v=f\times \lambda.........(2)

From equation (1) and (2) :

l=\dfrac{f^2\lambda^2m}{T}

l=\dfrac{(260\ Hz)^2\times (0.6\ m)^2\times 2.4\times 10^{-3}\ kg}{120\ N}

l = 0.486 m

So, the length of the string is 0.486 meters. Hence, this is the required solution.

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A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).
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Answer:

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b) The y coordinate of the third mass is -0.944 meters.

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Where:

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\vec r_{i} - Location of the i-th particle with respect to origin, measured in meters.

If we know that \vec r_{cm} = (-0.500\,m,-0.700\,m), m_{1} = 1\,kg, \vec r_{1} = (-1.20\,m, 0.500\,m), m_{2} = 4.50\,kg, \vec r_{2} = (0.600\,m, -0.750\,m) and m_{3} = 4\,kg, then the coordinates of the third particle are:

(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}

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(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)

(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

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