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Sound questions

  1. Nov 21, 2006 #1
    I have a few questions about my hw...

    1. Assume you visit a world where the atmosphere is made up of hydrogen. Your spacecraft can fly Mach 20 on Earth, as measured by sound at 5degreesC. On this planet, what is your Mach number if the speed of sound in hydrogen is 1267m/s?

    2. What is the Mach number for a shuttle if it starts re-entry after travelling around Earth in 1.495h at an orbital radius of 6.73 x 10^6m?
    Assume that the air temperature is -30degreesC.

    3. On a hot summer night (32degreesC) you are listening to a rock group in a stadium 350m away. Your friend is sitting in an air-conditioned room across the country, listening to broadcast radio. If the signal travels 30,000km up to a satellite that retransmits it, who hears the concert first?

    4. Determine the depth of water if an echo using sonar returs in 870ms and the speed of sound in water is 5300km/h

    I tried to do these questions but i really can't figure it out.... T.T

    Any help would be appreciated...^^
    Thank you
     
    Last edited: Nov 21, 2006
  2. jcsd
  3. Nov 21, 2006 #2
    Please provide your attempts on the questions and from there help can be provided!
     
  4. Nov 21, 2006 #3
    well the first two i don't have any idea... T.T
    but the last two i did do

    3. first, v= 3.0 x 10^8m/s + (0.6)(32deg.C)
    = 3.0 x 10^9m/s
    t=d/v
    = 350m/3.0x10^9m/s
    = 1.16 x 10^-7
    then, t=d/v
    = 3.0x10^7m/3.0x10^9m/s
    = 0.010s
    therefore, i hear the concert first...
    but i don't think it's right...
    the answer says that it takes 1.00s for for the first calculation

    4. t=0.87s d=vt
    v=1472.2m/s = (1472.2m/s)(0.87s)
    = 1280.8m
    the answer's 640m...
     
  5. Nov 21, 2006 #4
    1.
    Check out wikipedia.org for info about what mach numbers are. If speed of sound is used to define the mach number what happens if the speed of sound changes?

    2.
    No idea what 1.495h is :/

    3.
    The velocity regarding 350m is not the speed of light, it is the speed of sound... around 350m/s i believe. The second calculation seems correct.

    Yes 350/350 would give you 1s thus he would hear it first not you.

    4.
    The total distance traveled is 1280.8m yes but how do echoes work? They go there and back? thus...?
     
  6. Nov 21, 2006 #5
    1. so if the speed of sound increases, mach number increases and it would become supersonic..? and if it decreases, it becomes subsonic?

    2. I belive 1.495h means 1.495 hours...
     
  7. Nov 21, 2006 #6
    1.
    Erm I guess, doesn't really matter.

    Mach means V/Vsound

    Thus 20Mach would mean V=20*Vsound

    Now what's the new mach number in the hydrogen: mind that Vsound changed

    Mach number = V/Vsound(in hydrogen)

    2.
    havent got a clue
     
  8. Nov 21, 2006 #7
    um... i get a wrong answer...
    this is what i did...
    v= (332 +0.6(5degC))m/s
    = 335m/s
    Mach= v of hydrogen / v
    = 1267m/s / 335m/s
    = 3.78

    the answer's 5.29... what did i do wrong??
    and one more quick question...
    For the final answer for mach number how many sig figs do i have to have??? should it be 4 because 5degC has the least amt of digits so the final answer only has one sig fig? or...... i'm so confused with sig figs now...T.T
     
  9. Nov 21, 2006 #8
    From
    if you know the radius of the shuttle's orbit and the time it takes to complete an orbit, you can calculate its orbital speed. that's simple math.

    if you assume that upon re-entry that speed doesn't change, and at whatever altitude it "re-enters" the atmosphere, the speed of sound is the same as that of air at -30 degrees C, it's pretty straighforward to calculate how many times the speed of sound it's going at re-entry.

    does framing the question in those kinds of words and sentences help?

    see if you can get help from others here or from your teachers/instructors/professors on how to reword the given information so that the solutions are easier to frame.

    +af
     
  10. Nov 21, 2006 #9
    From
    Similarly, if the spacecraft files at Mach 20 "on earth" (over the earth), how many m/s or km/h is its ground speed?

    if it flies at that ground speed on the planet with the H2 atmosphere, what's its Mach speed?


    not so hard when you say it that way?

    again, +af
     
  11. Nov 21, 2006 #10
    i get a wrong answer...T.T
    this is what i did:
    v=d/t
    =6.73x10^6m/s / 5382s
    = 1260.46m's

    v=(332+0.6T)m/s
    = (332 + 0.6(-30degC))m/s
    = 314m/s

    Mach = 1250.46m/s / 314m/s
    = 4.0

    the answer's 25.....
     
  12. Nov 21, 2006 #11
    v= (332 +0.6(5degC))m/s
    = 335m/s
    Mach= v of hydrogen / v
    = 1267m/s / 335m/s
    = 3.78

    um... is this how you do it??????
    well i guess not cause the answer's wrong lol
    the answer's 5.29...

    and in this case how many sig figs would the final answer have???
     
  13. Nov 22, 2006 #12

    OlderDan

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    Mach number is the speed of the object (v) divided by the speed of sound (u)

    M = v/u = 20 in air where, by your calculation, u = 335m/s

    v = 20u = ___________

    In hydrogen, where the sound speed is u' = 1267m/s the mach number will be

    M' = v/u' = ____________
     
  14. Nov 22, 2006 #13
    thank you so much older dan
    i get it now! > . <
    but how many sig figs should my final answer have? it's confusing cause there were many steps involved in this...

    can u help me with #4 by any chance???
    i tried to do it, u'll see my posts above... but i get a wrong answer...T.T
     
  15. Nov 22, 2006 #14

    OlderDan

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    Not more than 3 on the significant figures. It's hard to say when a problem gives a number like 20 how they want you to take it. Do they mean within 5 of 20, or 20 +- .5, or 20.0 or ?????? When in doubt, I usually go with 3, and since you have your sound speed to 3, I would go with that.

    Problem #4 is just a distance, speed, time problem. The sound has to travel down and back, so the distance is twice the depth of the water, at the speed given in the problem, in the time specified. Just be careful with the units. Convert km/hr to km/sec or m/sec and ms (milliseconds) to seconds.
     
  16. Nov 22, 2006 #15
    oh oops i get #4 i meant #2 :redface:
     
  17. Nov 24, 2006 #16

    OlderDan

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    Can you find the speed from the given orbital radius and the period?
     
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