Respuesta :
Answer:
1.33 * 10⁻¹¹metres
Explanation:
Using the relationship to determine the amount by which the frequency of the line be shifted [tex]v = f\lambda[/tex]
v = velocity of the moving star = 7600m/s
f = frequency of the line spectrum = 5.71*10^14 Hz
from the formula;
[tex]\lambda = \frac{v}{f}\\ \lambda = \frac{7600}{5.71*10^{14} } \\\lambda = \frac{7.6*10^{3} }{5.71*10^{14} }\\\lambda = 1.33*10^{-11}m[/tex]
The frequency of the line will be shifted by 1.33 * 10⁻¹¹metres
Answer: 1.45x10^10 (or if your program already has x10^10 filled in and just wants the first part, your answer would just be 1.45)
This is the answer that my online school curriculum said was correct. Hope it helps anyone who needs it!