The center of mass is calculated using $$x_{cm} = \frac{\sum m_i x_i}{\sum m_i} = \frac{(2.0)(0) + (3.0)(4.0) + (1.0)(6.0)}{2.0 + 3.0 + 1.0} = \frac{0 + 12.0 + 6.0}{6.0} = \frac{18.0}{6.0} = 2.7\text{ m}$$. Choice A uses an incorrect calculation. Choice C represents the simple average of positions without mass weighting. Choice D is the position of the middle mass, not the center of mass.

Using the center of mass definition with $$m_1$$ at the origin: $$x_{cm} = \frac{m_2 d}{m_1 + m_2}$$ where $$d$$ is the separation. Since $$x_{cm} = 2.0\text{ m}$$: $$2.0 = \frac{8.0 \cdot d}{4.0 + 8.0} = \frac{8.0d}{12.0}$$. Solving: $$d = 3.0\text{ m}$$. Choice A equals the distance from $$m_1$$ to the center of mass. Choice C would be correct if the masses were equal. Choice D assumes an incorrect mass ratio calculation.

The uniform rod's center of mass is at $$\frac{L}{2}$$. Using the center of mass formula: $$x_{cm} = \frac{M \cdot \frac{L}{2} + m \cdot \frac{L}{4}}{M + m} = \frac{\frac{ML}{2} + \frac{mL}{4}}{M + m} = \frac{\frac{2ML + mL}{4}}{M + m} = \frac{L(2M + m)}{4(M + m)}$$. Choice B has an incorrect denominator factor. Choice C incorrectly weights the masses. Choice D uses an incorrect coefficient for the point mass.

The original system can be treated as a single particle of mass $$12\text{ kg}$$ at $$(3, 4)$$. The new center of mass is: $$x_{cm} = \frac{12 \cdot 3 + 4 \cdot 6}{12 + 4} = \frac{36 + 24}{16} = 3.75\text{ m}$$ and $$y_{cm} = \frac{12 \cdot 4 + 4 \cdot 1}{16} = \frac{48 + 4}{16} = 3.25\text{ m}$$. Choice A uses incorrect mass weighting. Choice C reverses the coordinate calculations. Choice D uses equal weighting instead of mass weighting.

Place the square with corners at $$(\pm a/2, \pm a/2)$$. Remove the mass at $$(a/2, a/2)$$. The remaining masses are at $$(-a/2, a/2)$$, $$(-a/2, -a/2)$$, and $$(a/2, -a/2)$$. The center of mass is at: $$x_{cm} = \frac{-a/2 - a/2 + a/2}{3} = -\frac{a}{6}$$ and $$y_{cm} = \frac{a/2 - a/2 - a/2}{3} = -\frac{a}{6}$$. The distance from origin is $$\sqrt{(a/6)^2 + (a/6)^2} = \frac{a\sqrt{2}}{6}$$. Choice A gives only one coordinate component. Choice B uses incorrect weighting. Choice D doubles the correct result.

The velocity of the center of mass is: $$\vec{v}_{cm} = \frac{\sum m_i \vec{v}_i}{\sum m_i} = \frac{2m(3\hat{i}) + m(-2\hat{i}) + 3m(1\hat{i})}{2m + m + 3m} = \frac{6m\hat{i} - 2m\hat{i} + 3m\hat{i}}{6m} = \frac{7m\hat{i}}{6m} = \frac{7}{6}\hat{i}\text{ m/s}$$. Choice A incorrectly weights the mass contributions. Choice B uses an arithmetic error in the numerator calculation. Choice D represents an incorrect total mass calculation.

For a uniform semicircular wire, using polar coordinates where $$y = R\sin\theta$$ and $$\theta$$ ranges from $$0$$ to $$\pi$$. The linear mass density is $$\lambda = M/(\pi R)$$. The $$y$$-coordinate of the center of mass is: $$y_{cm} = \frac{\int_0^\pi R\sin\theta \cdot \lambda \cdot R d\theta}{\int_0^\pi \lambda \cdot R d\theta} = \frac{\lambda R^2 \int_0^\pi \sin\theta d\theta}{\lambda R \cdot \pi} = \frac{R[-\cos\theta]_0^\pi}{\pi} = \frac{R(1-(-1))}{\pi} = \frac{2R}{\pi}$$. Choice A uses half the correct result. Choice C represents the geometric centroid of a semicircle incorrectly. Choice D includes an unnecessary $$\sqrt{2}$$ factor.

The velocity of the center of mass is calculated as: $$v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{3.0 \times 0 + 1.0 \times 8.0}{3.0 + 1.0} = \frac{0 + 8.0}{4.0} = 2.0\text{ m/s}$$. Choice A represents a quarter of the moving mass's velocity. Choice C uses incorrect mass ratios in the calculation. Choice D represents half of the moving mass's velocity without proper mass weighting.

For a uniform triangular plate, the center of mass is located at the centroid, which is the average of the vertex coordinates: $$x_{cm} = \frac{0 + 3 + 0}{3} = 1.0$$ and $$y_{cm} = \frac{0 + 0 + 4}{3} = \frac{4}{3} = 1.33$$. Choice B incorrectly uses the midpoint of the triangle's bounding rectangle. Choice C uses incorrect averaging for the $$y$$-coordinate. Choice D represents a weighted average that doesn't apply to uniform triangular plates.

For a uniform solid hemisphere, the center of mass is located at a distance of $$\frac{3R}{8}$$ from the flat face along the axis of symmetry. This can be derived using integration in spherical coordinates. When the hemisphere sits with its curved surface on the ground, the flat face is at height $$R$$ above the ground, but the center of mass is at $$\frac{3R}{8}$$ above the flat surface, which means it's at height $$R - \frac{3R}{8} = \frac{5R}{8}$$ above the ground. However, the question asks for the height above the flat surface, which is $$\frac{3R}{8}$$. Choice B gives an incorrect fraction. Choice C would be correct for a uniform semicircular disk. Choice D overestimates the center of mass location.