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The problem shown asks to find the equation relating the force \( F \), blade area \( A \), and wind velocity \( v \), given that the force varies jointly as the blade area \( A \) and the square of the wind velocity \( v \). The values provided are:

- \( F = 19.7 \, \text{lb} \)
- \( A = 3.74 \, \text{ft}^2 \)
- \( v = 31.6 \, \text{ft/s} \)

The general form of the joint variation is:

\[
F = k A v^2
\]

where \( k \) is the constant of proportionality. To find \( k \), substitute the given values of \( F \), \( A \), and \( v \) into the equation:

\[
19.7 = k \times 3.74 \times (31.6)^2
\]

First, calculate \( (31.6)^2 \):

\[
31.6^2 = 998.56
\]

Now substitute that into the equation:

\[
19.7 = k \times 3.74 \times 998.56
\]

\[
19.7 = k \times 3733.58
\]

Solve for \( k \):

\[
k = \frac{19.7}{3733.58} \approx 0.00527
\]

Thus, the equation relating \( F \), \( A \), and \( v \) is:

\[
F = 0.00527 A v^2
\]

The correct answer is option **C**: \( 0.00527 A v^2 \).

Would you like more details on any steps? Here are some follow-up questions to deepen your understanding:

1. What is the difference between direct and joint variation in equations?
2. How does squaring the velocity affect the proportionality of the force?
3. How can you use the concept of joint variation in other real-world situations?
4. What would happen to the force if both the area and velocity doubled?
5. How can the constant of proportionality \( k \) be interpreted physically?

**Tip**: Always check units in physical equations to ensure consistency throughout the problem!

The force F on the blade of a wind generator varies jointly as the blade area A and the square of the wind velocity v. Find the equation relating F, A, and v if F = 19.7 lb when A = 3.74 ft² and v = 31.6 ft/s.

Learn how to solve a joint variation problem involving the force on a wind generator blade. The force varies jointly as the blade area and the square of the wind velocity. Includes a step-by-step solution, using the equation F = kAv² with specific values. Suitable for college-level students.

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