Recent content by jdg

  1. J

    Speed of Sound in water given frequency

    I used the constant, 1493 m/s, which is given in my text for SoS in fresh water, it came out wrong
  2. J

    Speed of Sound in water given frequency

    so how do I find that speed with the given information?
  3. J

    Speed of Sound in water given frequency

    so it doesn't matter what the frequency is, the speed of sound remains the same in the water?
  4. J

    Speed of Sound in water given frequency

    the frequency won't change the speed in water?
  5. J

    Speed of Sound in water given frequency

    I would think that at a constant temperature the speed would be constant?
  6. J

    Speed of Sound in water given frequency

    I know it has something to do with the density of the medium, but I'm not sure what the equation is -the density of fresh water is 1000 kg/m3
  7. J

    Speed of Sound in water given frequency

    A 190 Hz sound traveling in fresh water has speed of? I'm not sure which equations to use, or where to start. I tried wavelength = V/f , with V being constant, 1493 m/sAny help would be greatly appreciated!
  8. J

    Continuity Equation problem

    Next question: A garden hose with internal diameter of 13.5 mm lies flat on a sidewalk while water is flowing in it at a speed of 6 m/s. A person happens to step on it at the very edge of the opening of the hose and decreases its internal diameter by a factor of 9 So D (1) = 0.0135m r (1) =...
  9. J

    Solve Bernoulli's Problem: Water Flow in Garden Hose

    Ok, for Q2, part 2 I did J = A(1)V(1) = A(2)V(2): So V2 = V1*(A1/A2) = 486 m/s Is this right? And for part 3 I did J = (A2)(V2) = 8.59e-4 m3/s
  10. J

    Solve Bernoulli's Problem: Water Flow in Garden Hose

    Next question: A garden hose with internal diameter of 13.5 mm lies flat on a sidewalk while water is flowing in it at a speed of 6 m/s. A person happens to step on it at the very edge of the opening of the hose and decreases its internal diameter by a factor of 9 So D (1) = 0.0135m r (1) =...
  11. J

    Solving a Buoyancy Problem with Carbon Steel

    for Q2, part 2 I did J = A(1)V(1) = A(2)V(2): So V2 = V1*(A1/A2) = 486 m/s Is this right? And for part 3 I did J = (A2)(V2) = 8.59e-4 m3/s
  12. J

    Solving a Buoyancy Problem with Carbon Steel

    ok, I got V = 0.003058... m = 23.975...
  13. J

    Solving a Buoyancy Problem with Carbon Steel

    Ok, I got: Fair= mg = 30 N m = dV = (7840)(3.901e-4) =3.06 kg V = Fb/dg = (30 N) / (7840*9.81) = 3.901 m3 Next question: A garden hose with internal diameter of 13.5 mm lies flat on a sidewalk while water is flowing in it at a speed of 6 m/s. A person happens...
  14. J

    Solving a Buoyancy Problem with Carbon Steel

    thanks a lot! you have time for one or 2 more?
  15. J

    Solving a Buoyancy Problem with Carbon Steel

    ok, so for weight in air it would be 30 N + 30 N= 60 N? And the mass would be m = Fair/g or m = (Fair-Fb)/g?
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