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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.

1. If p varies directly as q and the square of r and inversely as s, and p = 40 when q=5, r=4 and s=6.
    a. Find p when q = 8, r = 6 and s = 9.
    b. Find s when p=10, q=5, and r = 2.

2. The force of attraction F of a body varies directly as its mass m and inversely as the square of the distance d from the body. When m = 8 kilograms and d = 5 meters, F = 100 Newtons. Find F when m=2 kilograms and d = 5 meters.

Solve a direct and inverse variation problem involving proportionality constants, where p varies with q, r, and s, and F varies with mass and distance. This detailed solution is designed for students in Grades 9-11.

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