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The period of a 15.0 km radius star is indeed 2.9 s, the period has decreased to 2.4 s, calculate the new radius of the star

User Jaylon
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1 Answer

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Final answer:

The new radius of the star is approximately 18.79 km when the period decreases from 2.9 s to 2.4 s.

Step-by-step explanation:

The period of a rotating object is inversely proportional to the square of its radius. If the period of a 15.0 km radius star is initially 2.9 s and it decreases to 2.4 s, we can calculate the new radius using the period-radius relationship.

Let's assume that the initial period P_initial is equal to 2.9 s and the initial radius R_initial is equal to 15.0 km. Similarly, the final period P_final is equal to 2.4 s.

Using the formula P = kR^2, where k is a constant, we can set up the following proportion:

P_initial = kR_initial^2

P_final = kR_final^2

Dividing the two equations:

P_initial / P_final = (kR_initial^2) / (kR_final^2)

Simplifying further:

(P_initial / P_final) = (R_initial^2) / (R_final^2)

Substituting the given values:

(2.9 s / 2.4 s) = (15.0 km)^2 / (R_final^2)

Now, we can solve for R_final:

R_final^2 = (15.0 km)^2 * (2.4 s / 2.9 s)

R_final^2 = (15.0 km)^2 * 0.827586207

R_final^2 = 353.4482759 km^2

Taking the square root of both sides:

R_final = sqrt(353.4482759) km

R_final = 18.79 km

Therefore, the new radius of the star is approximately 18.79 km.

User Arun Chettoor
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