Given: - Frequency of the electromagnetic wave: \( f = 60 \, \text{MHz} = 60 \times 10^6 \, \text{Hz} \) - Speed of light in air: \( c = 3 \times 10^8 \, \text{m/s} \)
The wavelength \( \lambda \) of an electromagnetic wave is given by the formula:
\[ \lambda = \frac{c}{f} \]
Substituting the given values:
\[ \lambda = \frac{3 \times 10^8 \, \text{m/s}}{60 \times 10^6 \, \text{Hz}} \]
Simplifying:
\[ \lambda = \frac{3 \times 10^8}{60 \times 10^6} \, \text{m} \] \[ \lambda = \frac{3}{60} \times 10^2 \, \text{m} \] \[ \lambda = 5 \, \text{m} \]
The wavelength of the electromagnetic wave is \( 5 \, \text{m} \).
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).