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xz_007 [3.2K]
3 years ago
5

De la 2surse coerente aflate la distanta d=2 metri una de alta se propaga intrun mediu elastic unde cu viteza =340 m/s.la distan

ta L =4 metri de la mijlocul dintre surse perpendicular pe segmentul care le uneste a fost inregistrat un maxim de interferenta iar urmatorul la distanta de 1, 5 metri de la acesta pe linia paralela cu cea pe care se afla sursele.Care este frecventa de oscilatie a surselor?
Physics
1 answer:
Alja [10]3 years ago
5 0

<u>Explanation</u>:

The text is in Romanian language, however here's a clearer rendering which reads;

<em>"From 2 coherent sources at a distance d = 2 meters from each other, it propagates in an elastic medium where with velocity = 340 m/s. At a distance L = 4 meters from the middle between sources perpendicular to the segment joining them a maximum of interference and the next at a distance of 1.5 meters from it on the line parallel to the one on which the sources are located. What is the frequency of oscillation of the sources?."</em>

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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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The center of mass of a system of particles (\vec r_{cm}), measured in meters, is defined by this weighted average:

\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}} (1)

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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)

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

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