0.02_(KT)cdot 9.8M/c^2cdot 4Mcdot sin(30^circ )=(1)/(rho )

To solve the given equation, we need to find the value of ρ (rho).<br /><br />Given equation:<br />$0.02_{KT}\cdot 9.8M/c^{2}\cdot 4M\cdot sin(30^{\circ })=\frac {1}{\rho }$<br /><br />First, let's simplify the equation step by step:<br /><br />1. Convert $0.02_{KT}$ to $0.02$ (assuming it's a typo and should be $0.02$).<br />2. Calculate $sin(30^{\circ })$, which is $0.5$.<br />3. Multiply the constants together: $0.02 \cdot 9.8 \cdot 4 \cdot 0.5$.<br /><br />Now, let's perform the calculations:<br /><br />$0.02 \cdot 9.8 = 0.196$<br /><br />$0.196 \cdot 4 = 0.784$<br /><br />$0.784 \cdot 0.5 = 0.392$<br /><br />So, the equation becomes:<br /><br />$0.392 = \frac {1}{\rho }$<br /><br />To find ρ, we take the reciprocal of both sides:<br /><br />$\rho = \frac {1}{0.392}$<br /><br />Now, let's calculate the value of ρ:<br /><br />$\rho \approx 2.55$<br /><br />Therefore, the value of ρ is approximately $2.55$.