To solve part (b), we need to find the maximum revolutions per minute (denoted as $r$) of the propeller. The formula given is:

$d = 50 \left(\frac{h}{r^3}\right)^{1/5}$

Given:

• $h = 8500$ horsepower
• $d = 20$ feet

We need to solve for $r$.

Step-by-Step Solution:

1. Start with the formula:

   $20 = 50 \left(\frac{8500}{r^3}\right)^{1/5}$

2. Divide both sides by 50 to isolate the term involving $r$:

   $\frac{20}{50} = \left(\frac{8500}{r^3}\right)^{1/5}$

   Simplifying:

   $0.4 = \left(\frac{8500}{r^3}\right)^{1/5}$

3. Raise both sides to the power of 5 to eliminate the 1/5 exponent:

   $0.4^5 = \frac{8500}{r^3}$

   Simplifying $0.4^5$:

   $0.01024 = \frac{8500}{r^3}$

4. Multiply both sides by $r^3$ and divide by 0.01024 to solve for $r^3$:

   $r^3 = \frac{8500}{0.01024}$

5. Calculate:

   $r^3 = 830078.125$

6. Take the cube root of both sides to find $r$:

   $r = \sqrt[3]{830078.125}$

   Calculating the cube root:

   $r \approx 94.19 \text{ revolutions per minute (rpm)}$

Thus, the maximum revolutions per minute of the propeller should be approximately 94.19 rpm.

Would you like more details or have any questions?

Related Questions:

1. How would the propeller size change if the engine produced 9000 horsepower instead of 8500?
2. How does the power of the engine affect the propeller's diameter if rpm is constant?
3. What is the relationship between propeller size and revolutions per minute if horsepower is fixed?
4. How would doubling the revolutions per minute affect the propeller size if the engine power is constant?
5. If the propeller's diameter is increased to 25 feet, what should the new maximum rpm be?

Tip:

When solving equations involving exponents and roots, always isolate the term with the variable before applying powers or roots for a more straightforward solution!