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  1. M

    Air Bubbles and Gases - Please Help?

    V2 = V1*(T2 / T1)*(P1 / P2) V2 = (8.18 cm^3)*(289 K / 289 K)*(212 kPa / 101 kPa) V2 = (8.18 cm^3)*(2.099 kPa) V2 = 17.17 cm^3 Correct?
  2. M

    Air Bubbles and Gases - Please Help?

    T2 / t1 = 289 k / 289 k = 1 k?
  3. M

    Air Bubbles and Gases - Please Help?

    The temperature is a CONSTANT 16.0ºC...
  4. M

    Air Bubbles and Gases - Please Help?

    Pressure at the surface is 1 atm = 101 kPa. Pressure at 11 m = 2.09 atm = 212 kPa. V2 = V1*(T2 / T1)*(P1 / P2) V2 = (8.18 cm3)*(289 K)*(212 kPa / 101 kPa) V2 = 4962 cm3 That can't be right... What do I need to have everything converted to? Do I convert them to the following: T1 = 16.0ºC +...
  5. M

    Air Bubbles and Gases - Please Help?

    Homework Statement A scuba diver releases a 2.50-cm-diameter (spherical) bubble of air from a depth of 11 m in a lake. Assume the temperature is constant at 16.0ºC, and the air behaves as a perfect gas. How large is the bubble when it reaches the surface? Homework Equations V2 = V1*(T2...
  6. M

    Spring Constant and Oscillations - Please Help

    So I got 0.0182 with my calculator in radians... Is that correct?
  7. M

    Spring Constant and Oscillations - Please Help

    I changed my calculator to radians: x(t) = A sin (sqrt (k/m) * t) x(t) = (0.290) sin(sqrt (2.41 / 1.450) * 0.490T) x(t) = 0.145 m
  8. M

    Spring Constant and Oscillations - Please Help

    So I did: x(t) = A sin (sqrt (k/m) * t) x(t) = (0.290) sin(sqrt (2.41 / 1.450) * 0.490T) x(t) = (0.290) sin(1.29 * 0.490T) x(t) = (0.290) sin(0.6321T) x(t) = (0.290)(0.011) x(t) = -0.00319 m Does that seem correct to you?
  9. M

    Spring Constant and Oscillations - Please Help

    Homework Statement A 1.450 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.700 N to displace the glider to a new equilibrium position, x= 0.290 m. a.) Find the effective spring constant of the system. b.) The glider is...
  10. M

    Simple Pendulum - Maximum Speed

    I drew the diagram, but I'm not sure how to find the potential energy?
  11. M

    Simple Pendulum - Maximum Speed

    1. Homework Statement : A simple pendulum has a mass of 0.25 kg and a length of 1.0 m. It is displaced through an angle of 5.0° and then released. 1. What is the maximum speed? 2. What is the maximum angular acceleration? Answer: 0.855 rad/s2 3. What is the maximum restoring force? Answer...
  12. M

    Height of Blood Transfusion - Please Help?

    Okay. So to find pressure difference: P = 8 * η * L * Q / pi * r4 P = 8 * (0.004 Pa*s) * (0.0407 m) * (6.5 x 10-8 m3/s) / pi * (0.000194 m)4 P = 19023 Pa If that is correct, how do I find height from this?
  13. M

    Tomato, Salt Water & Olive Oil - Buoyancy

    Homework Statement 1. You put a tomato into a vessel of salty water - it floats. Now you pour olive oil into the same vessel. Which one of the following statements is true? a. More of the tomato will be above the surface of the water. b. The tomato will stay at the same height. c. Less of...
  14. M

    Flow Rate Problem -

    Could you please check out my question on Height of Blood Transfusion? Thanks.
  15. M

    Height of Blood Transfusion - Please Help?

    Homework Statement A patient is to be given a blood transfusion. The blood is to flow through a tube from a raised bottle to a needle inserted in the vein. The inside diameter of the 4.07 cm long needle is 0.388 mm and the required flow rate is 3.90 cm3 of blood per minute. How high should the...
  16. M

    Flow Rate Problem -

    You were right! Thank you for helping me out!
  17. M

    Flow Rate Problem -

    So if I recalculate, fixing my errors and taking out 0.5: t = V / Q = 8 * V * h * L / (pi * r^4 * Dp) t = 8 (0.006 m^3)*(0.2)*(0.12 m) / (pi * (0.026 m)^4 * (824 Pa)) t = 0.974 sec Still wrong.
  18. M

    Flow Rate Problem -

    You can calculate the pressure difference: Dp = rgL Dp = (700 kg/m^3)(9.81 m/s^2)(0.12 m) Dp = 824 Pa And the time it takes for the oil to flow through is the volume of the oil divided by the flow rate: t = V / Q = 8 * V * h * L / (pi * r^4 * Dp) t = 8 (0.006 m^3)*(0.2)*(0.012 m) / (pi *...
  19. M

    Flow Rate Problem -

    So I calculated the equation using pi: t = V / Q = 8 * V * h * L / (p * r^4 * Dp) t = 8 (0.006 m^3)*(0.2)*(0.012 m) / [ pi * (0.5)*(0.013 m)^4 * (82.4 Pa) ] t = 1.9492 sec But it is still incorrect... Any ideas?
  20. M

    Speed of Water Out Nozzle. Please Help?

    It is correct. Once again, thank you for spending the time to help me with this. I appreciate it tremendously.
  21. M

    Speed of Water Out Nozzle. Please Help?

    So if I fix the equation and recalculate: v = (0.019 m^3/s) / A = (0.019 m^3/s) / (3.14*(0.0217m)^2 / 4) = 51.4 Is the units m/s or m3/s?
  22. M

    Speed of Water Out Nozzle. Please Help?

    But do you really need radius if A = pi * d2 / 4?
  23. M

    Speed of Water Out Nozzle. Please Help?

    Yes, I wrote it down right. I will 12.8, but I'm not sure if it'll make a difference. But anyways, thank you for all of your help. =) I really appreciate it.
  24. M

    Speed of Water Out Nozzle. Please Help?

    It is wrong. I have no idea why. Ughhhh.
  25. M

    Speed of Water Out Nozzle. Please Help?

    A*v = 0.019 m^3/s v = (0.019 m^3/s) / A = (0.019 m^3/s) / (3.14*(0.0217m)^2) = 12.9 m/s Is that correct?
  26. M

    Speed of Water Out Nozzle. Please Help?

    So is v1 = v2 = 0.019 m^3/s?
  27. M

    Speed of Water Out Nozzle. Please Help?

    A*v describes the volume rate of low (m^3/s)... Right?
  28. M

    Speed of Water Out Nozzle. Please Help?

    It states that pressure is constant. But I'm still confused?
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