2.A hammer taps on the end of a 4.00-m-long metal bar at room temperature. A mic

Question

2.A hammer taps on the end of a 4.00-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pulses of sound, one that travels through the metal and one that travels through the air. The pulses are separated in time by 10.8ms .
What is the speed of sound in this metal?




11. mastering physics
A large solar panel on a spacecraft in Earth orbit produces 1.3kW of power when the panel is turned toward the sun.
What power would the solar cell produce if the spacecraft were in orbit around Saturn, 9.5 times as far from the sun?


12.An osprey

Explanation / Answer

Let t be the time for the sound in the metal

Therefore t + 0.0108 is the time for the sound in air


Assuming the speed of sound in air is 343 m/s we have

343*(t + 0.018) = 4.0 => t = 4.0/343 - 0.0108 = 8.61x10^-3s
Since they each travel the same distance we have

v*8.61x10^-3 = 4.0

So the speed in the metal = 4.0/8.61x10^-3 = 464.57/s

Note if you have a different room temperature the v in the metal will be slightly different

That is a good answer. 9.5 squared = 90.5; so a solar panel that produces a maximum of 90.5 watts in Earth orbit would produce one watt max in Saturn orbit. If that is 2 kilowatts; divide 2000 by 90.5 = 22.09 watts, maximum in Saturn orbit. Neil

ok... so this may sound a little silly, but it is a great way to think about the doppler effect

so you are standing still

the bird is flying to you

imagine the 'sound' being emitted by the bird is actually a stream of pellets released every T seconds from the bird's mouth. T is the period (1/f) of the frequency.

if the bird is stationary, the pellets will be separated by a distance vT, ie after being released a pellet will travel a distance vT before the next one is released. vT is the wavelength of the sound.

if the bird is moving towards you, the distance between pellets will be SMALLER. In particular, if a pellet is released at a time 0, it will travel a distance vT before the next one is released, but in that time the bird will have traveled a distance uT where u is the velocity of the bird. thus, the apparent distance between pellets is (v-u)T- the apparent wavelength is (v-u)T.

thus, the apparent frequency is v divided by the apparent wavelength, which is

v/((v-u)T) = f v/(v-u)

where f is the original frequency

so if we call f' the frequency you hear, and solve for u

f/f' = (v-u)/v

u=v(1-f/f') = 343(1-2200/2300) = 14.9 m/s

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