- #1
buttterfly41
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A flower pot is knocked off a balcony 21.6 m above the sidewalk and falls toward an unsuspecting 1.79 m-tall man who is standing below. How close to the sidewalk can the flower pot fall before it is too late for a shouted warning from the balcony to reach the man in time? Assume that the man below requires 0.300 s to respond to the warning.
So i (21.6-1.79)m / 343m/s = .05776s + .3s = .35776s to yell and have the man react
then, 21.6m= 1/2 * 9.8m/s2 *T^2 ... T= 2.0996 - .35776s = 1.74181s for latest time to wait before yelling,
so i thought the answer should be: 1/2 * 9.8m/s2 * 1.7418^2 = 14.86m down, so 5.74m from the ground... but that is not correct :(
any ideas where i went wrong?
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Also, A sound wave in air has a pressure amplitude equal to 3.94 X10^-3 Pa. Calculate the displacement amplitude of the wave at a frequency of 10.5 kHz.
(Note: Use the following values, as needed. The equilibrium density of air is p= 1.20 kg/m3; the speed of sound in air is v = 343 m/s. Pressure variations P are measured relative to atmospheric pressure, 1.013 x10^5 Pa.)
so, i thought i would use the equation deltaPmax= pwvsmax
and then i plugged the numbers in (w=2pif=65973)... 1.013E5 = 1.2*65973*343*smax
smax= .00373m ... but again, wrong... so i don't know where to go from here
any help would be greatly appreciated. Thanks
So i (21.6-1.79)m / 343m/s = .05776s + .3s = .35776s to yell and have the man react
then, 21.6m= 1/2 * 9.8m/s2 *T^2 ... T= 2.0996 - .35776s = 1.74181s for latest time to wait before yelling,
so i thought the answer should be: 1/2 * 9.8m/s2 * 1.7418^2 = 14.86m down, so 5.74m from the ground... but that is not correct :(
any ideas where i went wrong?
----------
Also, A sound wave in air has a pressure amplitude equal to 3.94 X10^-3 Pa. Calculate the displacement amplitude of the wave at a frequency of 10.5 kHz.
(Note: Use the following values, as needed. The equilibrium density of air is p= 1.20 kg/m3; the speed of sound in air is v = 343 m/s. Pressure variations P are measured relative to atmospheric pressure, 1.013 x10^5 Pa.)
so, i thought i would use the equation deltaPmax= pwvsmax
and then i plugged the numbers in (w=2pif=65973)... 1.013E5 = 1.2*65973*343*smax
smax= .00373m ... but again, wrong... so i don't know where to go from here
any help would be greatly appreciated. Thanks