The mass of the plate is calculated in three sections:
\(m1 = σ × 5 = 10 kg,\)
\(m2 = σ × 1 = 2 kg,\)
\(m3 = σ × 6 = 12 kg.\)
Using the coordinates of each center of mass, we calculate the combined center of mass:
\[m_1x_1 + m_2x_2 = m_3x_3.\]
\[10 \cdot 1.5 + 2 \cdot (1.5) = 12 \cdot x_1 \implies x_1 = 1.5 \, \text{cm}.\]
Similarly:
\[m_1y_1 + m_2y_2 = m_3y_3.\]
\[10 \cdot 1 + 2 \cdot (1.5) = 12 \cdot y_1 \implies y_1 = 0.9 \, \text{cm}.\]
The ratio of $x_1$ to $y_1$ is:
\[\frac{x_1}{y_1} = \frac{1.5}{0.9} = \frac{15}{9}.\]
Thus:
\[n = 15.\]
Final Answer: $n = 15$.
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).