[math] m = 0.1g = 0.1 \times 10^{-3} kg [/math]

[math] \Delta t = 0.1 s [/math]

 

PE of ball = [math] (0.1 \times 10^{-3} kg)(9.81 ms^{-2})(1m) [/math]

 

but only 0.05% of it is converted into a sound wave so :

 

Energy (E) = [math] (0.0005)(0.1 \times 10^{-3} kg)(9.81 ms^{-2})(1m) [/math]

 

Now Power is Energy per unit time, so Power (P) is:

 

[math] E = \frac{(0.0005)(0.1 \times 10^{-3} kg)(9.81 ms^{-2})(1m)}{0.1s} [/math]

 

Intensity is Power per unit area (since it can be approimated to a point source of sound the area is the surface area of a sphere, [math] 4 \pi r^2 [/math], so Intensity is :

 

[math] I = \frac{(0.0005)(0.1 \times 10^{-3} kg)(9.81 ms^{-2})(1m)}{(0.1s)(4 \pi r^2)} [/math]

 

Since we are given that the lowest audible intensity is [math] 10^{-11} Wm^{-2} [/math] then we set I = [math] 10^{-11} Wm^{-2} [/math]

so we have:

 

[math] \frac{(0.0005)(0.1 \times 10^{-3} kg)(9.81 ms^{-2})(1m)}{(0.1s)(4 \pi r^2)} = 10^{-11} Wm^{-2} [/math]

 

So then rearranging and solving for r gives:

 

r = 200 m ( to 1 sig. fig.)

 

now for part (b) we just set I = [math] 10^{-8} Wm^{-2} [/math] instead, so we have :

 

[math] \frac{(0.0005)(0.1 \times 10^{-3} kg)(9.81 ms^{-2})(1m)}{(0.1s)(4 \pi r^2)} = 10^{-8} Wm^{-2} [/math]

 

now rearraging for r gives :

 

r = 6 m (to 1 sig. fig.)

oh yeah, i just wanted to know if i had done things correctly? :S

hello?

Your approach seems good enough ^^. Should be okay.

right ok, thanks