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: \[ W = F \cdot (5v) = Fv. \]
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: