The energy equivalent of a mass is given by Einstein’s equation:
\( E = mc^2, \)
where:
Step 1: Substitute values into \( E = mc^2 \)
\( E = (10^{-3}) \times (3 \times 10^8)^2 \, \text{J}. \)
Simplify:
\( E = (10^{-3}) \times (9 \times 10^{16}) \, \text{J}. \)
\( E = 9 \times 10^{13} \, \text{J}. \)
Step 2: Convert energy to electron volts (eV)
Using the conversion \( 1 \, \text{J} = 6.241 \times 10^{18} \, \text{eV}: \)
\( E = (9 \times 10^{13}) \times (6.241 \times 10^{18}) \, \text{eV}. \)
\( E = 56.169 \times 10^{31} \, \text{eV}. \)
Convert to MeV (\( 1 \, \text{MeV} = 10^6 \, \text{eV} \)):
\( E = 56.169 \times 10^{25} \, \text{MeV}. \)
Approximate:
\( E \approx 5.6 \times 10^{26} \, \text{MeV}. \)
Final Answer: \( 5.6 \times 10^{26} \, \text{MeV}. \)
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is: