56/3 °C
42/3 °C
This problem involves Newton's Law of Cooling, which can be expressed as: \[ \frac{dT}{dt} = -k(T - T_{\text{air}}), \] where \( T \) is the temperature of the body, \( T_{\text{air}} \) is the ambient temperature (temperature of the air), and \( k \) is a constant. The temperature changes from \( 40^\circ \text{C} \) to \( 24^\circ \text{C} \) in 4 minutes. Using Newton's Law of Cooling, we can compute the constant \( k \) and then apply it to determine the temperature change in the next 4 minutes. Based on the given information and applying the necessary calculations, the temperature after 4 more minutes is:
Final Answer:
A string of length \( L \) is fixed at one end and carries a mass of \( M \) at the other end. The mass makes \( \frac{3}{\pi} \) rotations per second about the vertical axis passing through the end of the string as shown. The tension in the string is ________________ ML.
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: