The work done by a constant force \( F \) is given by the formula: \[ W = F \cdot d, \] where \( d \) is the displacement of the block. If the block is moving with velocity \( v \), after time \( t = 5 \) seconds, the displacement will be \( d = v \cdot 5 \).
Thus, the work done is: \[ W = F \cdot (5v) = Fv. \]
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).