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2 years ago

Bx = -1.33 m and By = 2.81 m Find the magnitude of the vector.

Physics

1 answer:

NISA [10]2 years ago

3 0

Answer:

Explanation:

The formula for the magnitude of a vector is

$B_{mag}=\sqrt{(-1.33)^2+(2.81)^2}$ and then round to the hundredths place:

3.11 m. Since we are in Q2, we can also find the direction of this vector:

$tan^{-1}(\frac{2.81}{-1.33})=-64.7$ but since we are in Q2, we add 180 degrees to the result, getting the angle to be 115.3

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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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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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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:

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

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

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

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

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

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