The initial kinetic energy of the bullet is:
\[ K_i = \frac{1}{2} m (100)^2 \]
The final kinetic energy of the bullet is:
\[ K_f = \frac{1}{2} m (40)^2 \]
The percentage loss in kinetic energy is given by:
\[ \% \text{loss} = \frac{|K_f - K_i|}{K_i} \times 100 \]
Substituting the expressions for \( K_i \) and \( K_f \):
\[ \% \text{loss} = \frac{\left| \frac{1}{2} m (40)^2 - \frac{1}{2} m (100)^2 \right|}{\frac{1}{2} m (100)^2} \times 100 \]
Simplify:
\[ \% \text{loss} = \frac{|1600 - 10000|}{10000} \times 100 \]
\[ \% \text{loss} = \frac{8400}{10000} \times 100 = 84\% \]
Final Answer: 84% (Option 4)
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).