The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between \( t = 0 \) to \( t = 4s \)? 
The distance covered by an object is given by the area under the velocity-time graph.
Here, the graph shows a combination of trapezoidal and rectangular areas.
The total area under the graph between \( t = 0 \) and \( t = 4 \) represents the distance traveled.
By calculating the area from the graph: \[ {Distance} = {Area under the graph} = 13 \, {m} \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)