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Simora [160]
3 years ago
10

Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (

0.0, 3.2) m, and 4.0 kg at (2.5, 0.0) m. Where should a fourth object of 8.3 kg be placed so that the center of gravity of the four-object arrangement will be at (0.0, 0.0) m
Physics
1 answer:
Sauron [17]3 years ago
5 0

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}

10 + 8.3 x = 0

x = -1.20 m

Similarly for y direction we have

y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}

9.28 + 8.3 x = 0

x = -1.12 m

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Answer:

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Substituting,

C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F

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U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J

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3 years ago
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Look at the attached graphic:

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