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Thepotemich [5.8K]
3 years ago
12

ANYONE help me with this

Physics
1 answer:
ycow [4]3 years ago
8 0

Displacement is simply the change in position, or the difference in the final and initial positions:

\Delta d = d_f - d_i

Then

(a) ∆<em>d</em> = 5 m - 0 m = 5 m

(b) ∆<em>d</em> = 1 m - (-2 m) = 1 m + 2 m = 3 m

(c) ∆<em>d</em> = 2 m - (-2 m) = 2 m + 2 m = 4 m

(d) ∆<em>d</em> = 6 m - 2 m = 4 m

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X_{cm} = \frac{m_1x_1+m_2x_2....m_nX_n}{m_1+m_2...m_n}

where X_{cm} is the location of the center of gravity in the axis x, m_i is the mass of the object i and x_i the first coordinate of center of gravity of object i.

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0 = \frac{(5kg)(0)+(3kg)(0)+(4kg)(3)+(8kg)x_4}{5kg+3kg+4kg+8kg}

Where x_4 is the first coordinate of the center of gravity for the fourth object.

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At the same way:

Y_{cm} = \frac{m_1y_1+m_2y_2....m_ny_n}{m_1+m_2...m_n}

where Y_{cm} is the location of the center of gravity in the axis y, m_i is the mass of the object i and y_i the second coordinate of center of gravity of object i. replacing values we get:

Y_{cm} = \frac{(5kg)(0)+(3kg)(4)+(4kg)(0)+(8kg)y_4}{5+3+4+8}

Where y_4 is the second coordinate of the center of gravity for the fourth object.

solving for y_4:

y_4 = -1.5m

It means that the object of mass 8kg have to be placed in the  

coordinates (-1.5,-1.5) m.

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