Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago

How long will it take?

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  • NCS
    Lv 7
    2 months ago
    Favorite Answer

    a) p = ρgh = 1000kg/m³ * 9.8m/s² * 0.060m = 588 Pa

    b) By Bernoulli,

    p = ½ρv² → 588 Pa = ½*1000kg/m³*v² → v = 1.08 m/s

    c) Q = vA = 1.08m/s * (π/4)(0.0006m)²

    Q = 3.07e-7 m³/s = 3.07e-4 L/s = 0.307 mL/s

    _______

    d) t = V / Q = 500mL / 0.307mL/s = 1630 s

    e) We should factor in at least a couple of things. We can apply pressure (through body mechanics) beyond that produced by the static pressure of the fluid. Also, the urethra probably expands somewhat while the urine is passing through it (less than 1 mm seems quite small, but 5 mm is much too large).

  • Anonymous
    2 months ago

    Thanks very much for the answer but I do not understand part d and e.

    In d, why was 0.5 was multiplied to 10^4 then divided to 3 instead of 0.0003.

    In e, why was 0.5 was multiplied to 10^3 then divided to 21 instead of 0.021.

    Thanks

  • oubaas
    Lv 7
    2 months ago

    When your urinary bladder is full, the bladder pressure can reach up to 60 𝑚𝑚 of fresh water .

    a. What is this pressure p in units of Pascal?

    p = 60/1000 m * 1000 kg/m^3 * 9.806 N/kg = 588 Pa

    b. Assuming that there is no height difference between your urinary bladder and where your urine comes out, calculate the speed at which your urine comes out. The density of urine is 1030 𝑘𝑔/𝑚3.

    588 = 1030/2*V^2

    V = √ 588/515 = 1.07 m/sec

    c. If the diameter of a urethra is 0.6 𝑚𝑚, estimate the volume flow rate of urine as it comes out in units of liters per second.

    Flow = 1.07*10 dm/sec * 0.7854*0.006^2 dm^2 = 0.00030 dm^3/sec ( lps)

    d. If a full bladder constitutes 500 𝑚𝐿 of urine, how long t will it take you to remove all of the urine from your bladder?

    t = 0.5*10^4/3 = 1.660 sec

    e. The answer in part (d) is not a realistic time for peeing? What should we factor in in order to make it more realistic?

    Totally unrealistic ; the real time should in the range of half minute

    the real average diameter of a urethra is 5 𝑚𝑚, therefore :

    Flow = 1.07*10 dm/sec * 0.7854*0.05^2 dm^2 = 0.021 dm^3/sec ( lps)

    t = 0.5*10^3/21 = 25 sec

    note : 0.7854 = PI/4 

    comment 1 = dividind by 0,0003 means multiplying by 10^4 and dividing by 3

    comment 2 = dividind by 0,021 means multiplying by 10^3 and dividing by 21 

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