A girl stands at a distance of 170 m from a wall and claps her hands rhythmically. She makes 60 claps in 1 minute. She then moves to a new position and makes 45 claps in 1 minute. Find the speed of so

The girl makes 60 claps in 1 minute (60 seconds). Since each clap coincides with the echo of the one before, the time interval between successive claps is the time it takes for the sound to travel to the wall and back.

$$t = \frac{\text{Total time}}{\text{Number of claps}} = \frac{60 \text{ s}}{60} = 1 \text{ s}$$

The distance to the wall is $D = 170 \text{ m}$. The sound travels $2D$ for an echo.
The speed of sound $v$ is given by the formula:

$$v = \frac{2D}{t}$$

Substitute the values:

$$v = \frac{2 \times 170 \text{ m}}{1 \text{ s}}$$ $$v = 340 \text{ m/s}$$

The speed of sound in air is $\boxed{\text{340 m/s}}$.

The girl now makes 45 claps in 1 minute (60 seconds).

$$t' = \frac{\text{Total time}}{\text{Number of claps}} = \frac{60 \text{ s}}{45} = \frac{4}{3} \text{ s}$$

Using the speed of sound calculated in part (a), $v = 340 \text{ m/s}$, and the new time interval $t'$, the new distance $D'$ is:

$$D' = \frac{v \times t'}{2}$$

Substitute the values:

$$D' = \frac{340 \text{ m/s} \times \frac{4}{3} \text{ s}}{2}$$ $$D' = \frac{340 \times 4}{3 \times 2} \text{ m}$$ $$D' = \frac{1360}{6} \text{ m}$$ $$D' = \frac{680}{3} \text{ m}$$ $$D' \approx 226.67 \text{ m}$$

Her new distance from the wall is $\boxed{\text{226.67 m}}$.