Let’s analyze the forces acting on each block.
For the system as a whole (masses \(M_1\), \(M_2\), and \(M_3\) together) moving upwards with an acceleration \(a = 2 \, \mathrm{m/s^2}\):
Total mass, \(M = M_1 + M_2 + M_3 = 4 + 6 + 10 = 20 \, \mathrm{kg}.\)
Total weight, \(W = Mg = 20 \times 10 = 200 \, \mathrm{N}\)
Since the entire system is accelerating upwards, the net force \(F\) required to produce this acceleration is given by:
\(F = Ma = 20 \times 2 = 40 \, \mathrm{N}\)
Thus, the tension \(T_1\) in rope 1 must support both the weight and the additional force required for acceleration:
\(T_1 = W + F = 200 + 40 = 240 \, \mathrm{N}\)
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).