To determine the wavelength of an electromagnetic wave with a given frequency, we can use the fundamental relationship between the speed of light, frequency, and wavelength. The formula is given by:
\(c = \lambda \cdot f\)
Where:
Given:
We need to find the wavelength \(\lambda\) in meters.
Using the formula, we rearrange for \(\lambda\):
\(\lambda = \frac{c}{f}\)
Substitute the given values:
\(\lambda = \frac{3 \times 10^8}{60 \times 10^6}\)
Calculate \(\lambda\):
\(\lambda = \frac{3 \times 10^8}{60 \times 10^6} = \frac{3}{60} \times 10^2 = 0.05 \times 10^2 = 5\) meters.
Thus, the wavelength of the electromagnetic wave is 5 meters. Therefore, the correct answer is:
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 \( 5 \, \text{m} \).
A laser beam has intensity of $4.0\times10^{14}\ \text{W/m}^2$. The amplitude of magnetic field associated with the beam is ______ T. (Take $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/\text{N m}^2$ and $c=3\times10^8\ \text{m/s}$)
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.