Question

\[ \frac{m_{S}}{m_{P}}=\frac{\text { density }_{S}}{\text { density }_{P}} \times \frac{\text { thenness }_{S}}{\text { thickness }_{P}} \times \frac{\text { area }_{S}}{\text { area }_{P}} \] Home - Google Drive *3 Figure 9-36 shows a slab with dimensions \( d_{1}=11.0 \mathrm{~cm}, d_{2}= \) 2.80 cm , and \( d_{3}=13.0 \mathrm{~cm} \). Half the slab consists of aluminum (density \( =2.70 \mathrm{~g} / \mathrm{cm}^{3} \) ) and half consists of iron (density \( =7.85 \mathrm{~g} / \mathrm{cm}^{3} \) ). What are (a) the \( x \) coordinate, (b) the \( y \) coordinate, and (c) the \( z \) coordinate of the slab's center of mass? \[ \begin{array}{l} \text { Iron com }\left(-\frac{d_{3}}{2}, \frac{d_{1}}{2}, \frac{d_{2}}{2}\right) \\ \text { Aluminum com }\left(-\frac{d_{3}}{2}, d_{1}+\frac{d_{1}}{2}, \frac{d_{2}}{2}\right) \\ X_{\text {com }}=-\frac{d_{3}}{2}=\frac{-13.0 \mathrm{~cm}}{2}=-6.5 \mathrm{~cm} \\ z_{\text {Com }}=\frac{d_{2}}{2}=\frac{2.8 \mathrm{~cm}}{2}=1.4 \mathrm{~cm} \\ y_{\text {Com }}=\frac{m_{A L} y_{\text {Com,AL }}+m_{\mathrm{Fe}} y_{\text {Com, } \mathrm{Fe}}}{m_{A L}+m_{\mathrm{Fc}}} \end{array} \]

          \[
\frac{m_{S}}{m_{P}}=\frac{\text { density }_{S}}{\text { density }_{P}} \times \frac{\text { thenness }_{S}}{\text { thickness }_{P}} \times \frac{\text { area }_{S}}{\text { area }_{P}}
\]
Home - Google Drive
*3 Figure 9-36 shows a slab with dimensions \( d_{1}=11.0 \mathrm{~cm}, d_{2}= \) 2.80 cm , and \( d_{3}=13.0 \mathrm{~cm} \). Half the slab consists of aluminum (density \( =2.70 \mathrm{~g} / \mathrm{cm}^{3} \) ) and half consists of iron (density \( =7.85 \mathrm{~g} / \mathrm{cm}^{3} \) ). What are (a) the \( x \) coordinate, (b) the \( y \) coordinate, and (c) the \( z \) coordinate of the slab's center of mass?
\[
\begin{array}{l}
\text { Iron com }\left(-\frac{d_{3}}{2}, \frac{d_{1}}{2}, \frac{d_{2}}{2}\right) \\
\text { Aluminum com }\left(-\frac{d_{3}}{2}, d_{1}+\frac{d_{1}}{2}, \frac{d_{2}}{2}\right) \\
X_{\text {com }}=-\frac{d_{3}}{2}=\frac{-13.0 \mathrm{~cm}}{2}=-6.5 \mathrm{~cm} \\
z_{\text {Com }}=\frac{d_{2}}{2}=\frac{2.8 \mathrm{~cm}}{2}=1.4 \mathrm{~cm} \\
y_{\text {Com }}=\frac{m_{A L} y_{\text {Com,AL }}+m_{\mathrm{Fe}} y_{\text {Com, } \mathrm{Fe}}}{m_{A L}+m_{\mathrm{Fc}}}
\end{array}
\]

(mS)/(mP)=( density S)/( density P)×( thenness S)/( thickness P)×( area S)/( area P)

Home - Google Drive
*3 Figure 9-36 shows a slab with dimensions d1=11.0 cm, d2= 2.80 cm , and d3=13.0 cm. Half the slab consists of aluminum (density =2.70 g / cm^3 ) and half consists of iron (density =7.85 g / cm^3 ). What are (a) the x coordinate, (b) the y coordinate, and (c) the z coordinate of the slab's center of mass?

Iron com (-(d3)/(2), (d1)/(2), (d2)/(2))
Aluminum com (-(d3)/(2), d1+(d1)/(2), (d2)/(2))

Xcom =-(d3)/(2)=(-13.0 cm)/(2)=-6.5 cm

zCom =(d2)/(2)=(2.8 cm)/(2)=1.4 cm

yCom =(mA L yCom,AL +mFe yCom, Fe)/(mA L+mFc)

Added by Lisa B.

Question

Please give Ace some feedback