Let's solve each question step-by-step.

Problem 1:

We know that $p$ varies directly as $q$ and the square of $r$, and inversely as $s$. This can be represented by the equation: $p = k \frac{q \cdot r^2}{s}$ where $k$ is the constant of proportionality.

Given:

When $p = 40$, $q = 5$, $r = 4$, and $s = 6$: $40 = k \frac{5 \cdot 4^2}{6}$

1. Solve for $k$: $40 = k \frac{5 \cdot 16}{6}$ $40 = k \frac{80}{6}$ $40 = \frac{80k}{6}$ Multiply both sides by 6: $240 = 80k$ $k = \frac{240}{80} = 3$

So, the equation becomes: $p = 3 \frac{q \cdot r^2}{s}$

Part (a): Find $p$ when $q = 8$, $r = 6$, and $s = 9$.

$p = 3 \frac{8 \cdot 6^2}{9}$ $p = 3 \frac{8 \cdot 36}{9}$ $p = 3 \frac{288}{9}$ $p = 3 \cdot 32 = 96$

So, $p = 96$.

Part (b): Find $s$ when $p = 10$, $q = 5$, and $r = 2$.

$10 = 3 \frac{5 \cdot 2^2}{s}$ $10 = 3 \frac{5 \cdot 4}{s}$ $10 = 3 \frac{20}{s}$ Multiply both sides by $s$: $10s = 60$ $s = \frac{60}{10} = 6$

So, $s = 6$.

Problem 2:

The force $F$ varies directly as $m$ and inversely as $d^2$. This gives us: $F = k \frac{m}{d^2}$

Given:

When $F = 100$, $m = 8$, and $d = 5$: $100 = k \frac{8}{5^2}$ $100 = k \frac{8}{25}$

1. Solve for $k$: $k = \frac{100 \cdot 25}{8} = \frac{2500}{8} = 312.5$

So, the equation becomes: $F = 312.5 \frac{m}{d^2}$

Find $F$ when $m = 2$ and $d = 5$.

$F = 312.5 \frac{2}{5^2}$ $F = 312.5 \frac{2}{25}$ $F = 312.5 \cdot 0.08 = 25$

So, $F = 25$ Newtons.

Would you like more details on any part of the solution?

Follow-up Questions:

1. How would the value of $p$ change if $s$ increased in Problem 1?
2. What would happen to $F$ in Problem 2 if $d$ were halved?
3. In Problem 1, how would the constant $k$ change if the units of $p$, $q$, $r$, and $s$ were different?
4. What if both $m$ and $d$ doubled in Problem 2, what would be the new $F$?
5. Can you derive a formula for $s$ in terms of $p$, $q$, $r$, and $k$ from Problem 1?

Tip:

In problems involving direct and inverse variation, isolate the constant $k$ first; it simplifies the calculations for different scenarios.