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  1. M

    Lenses/Images - Find final image

    Lenses/Images -- Find final image Two converging lenses, each having a focal length equal to 8 cm, are separated by 35 cm. An object is 30 cm to the left of the first lens. (a) Find the position of the final image using both a ray diagram and the thin-lens equation (c) What is the overall...
  2. M

    Heat/First Law of Thermodynamics/Calorimetry - Have work done

    Heat/First Law of Thermodynamics/Calorimetry -- Have work done A calorimeter of negligible mass contains 920 g of water at 303 K and 48 g of ice at 273 K. Find the final temperature T. °C Solve the same problem if the mass of ice is 460 g. °C Formula Ive used Mice * Lf + Mice * Cwater *...
  3. M

    Simple Harmonic Motion of an oscillating particle

    The period of an oscillating particle is 56 s, and its amplitude is 18 cm. At t = 0, it is at its equilibrium position. Find the distances traveled during these intervals. (a) t = 0 to t = 14 s cm (b) t = 14 s to t = 28 s cm (c) t = 0 to t = 7 s cm (d) t = 7 s to t = 14 s cm x =...
  4. M

    Bernoulli's Equation on water hose

    Water flows at 0.67 m/s through a 3.0 cm diameter hose that terminates in a 0.35 cm diameter nozzle. Assume laminar non-viscous steady-state flow. (a) At what speed does the water pass through the nozzle? 49.22 m/s (b) If the pump at one end of the hose and the nozzle at the other end are...
  5. M

    Fluids - pressure

    where/how would the density of mercury come in? Im not too sure about this. They want the height in cm. 10mmHg = 1.03 kg/m^3 * 9.81N/kg * h
  6. M

    Fluids - pressure

    Oh so the density is 1.03, then g would be 9.81m/s then. P = P0 + density * g * h 10mmHg = 1.03 * 9.81m/s * h and solve for h?
  7. M

    Fluids - pressure

    Fluids -- pressure Blood plasma flows from a bag through a tube into a patient's vein, where the blood pressure is 10 mmHg. The specific gravity of blood plasma at 37°C is 1.03. What is the minimum elevation the bag must have so the plasma flows into the vein? Prof gave this problem for us...
  8. M

    Gravity - maximum height

    h = V^2R^2/(2*u - V^2*R) h = 13.69 * 40960000 / (800000 - 87616) h = 560742400 / 712384 h = 787.1350283 km h = .7871350283 m Do I add this to the Radius of the earth?
  9. M

    Gravity - maximum height

    well the textbook has the same problem explained but different velocity and it uses that equations and same steps I posted. But I don't know what I did wrong mathematically to give me a negative height. I was wondering if someone can work it out and see if they get the same answer or diff.
  10. M

    Gravity - maximum height

    This is the final equation: 1/radius_f = -Vi^2/2*G*M_e + 1/R_e then I solve for radius_f then I divide it by 1 then I subtract it from the earth's radius to get the height but it gives me a negative height.
  11. M

    Gravity - maximum height

    Gravity -- maximum height An object is projected upward from the surface of the Earth with an initial speed of 3.7 km/s. Find the maximum height it reaches. m Ive used the Energy conservation Kf + Uf = Ki + Ui 1/2 mvf^2 - G*M_e*m/rf = 1/2 mvi^2 - G*M_e*m/ri = 0 - G*M_e*m/rf = 1/2 mvi^2 -...
  12. M

    Gravity Problem - satellite

    I found the speed from part b 2.15 km/s circumference of the orbit is 2Pi*(86.37 x 10^6) so time = 2Pi*(86.37 x 10^6) / 2.15 km/s
  13. M

    Gravity Problem - satellite

    Yes, it makes sense, thanks for explaining. And I got what r is now that you related it as a circle. For part C: What you said earlier time = distance/speed where distance is the orbits circumference/distance covered It would be 2Pi*(86.37 x 10^6) / 2.15km I get a big answer of 252528.6976s...
  14. M

    Gravity Problem - satellite

    Ok, I got the mass of the earth 5.98 × 10^24 kilograms from the textbook. So the raidus of the Earth is 6.37 x 10^6m which my textbook says. So I add 8.00 x 10^7 m to it? So is 86,370,000 which is 86.37 x 10^6 gravitational Force = G*m1*m2 / r^2 G = 6.67 x 10^-11 Nm^2/kg^2 F = (6.67 x...
  15. M

    Gravity Problem - satellite

    Gravity Problem -- satellite A satellite with a mass of 250 kg moves in a circular orbit 8.00 x 10^7 m above the Earth's surface. (a) What is the gravitational force on the satellite? N (b) What is the speed of the satellite? km/s (c) What is the period of the satellite? h Relevant Equations...
  16. M

    Conservation of energy of a pendulum

    An 80cm long pendulum with a .60kg bob is released from rest at an initial angle of Theta with the vertical. At the bottom of the swing, the speed of the bob is 2.8m/s. a)what is Theta? b)What angle does the pendulum makes with the vertical when the speed of the bob is 1.4m/s? Revelant...
  17. M

    Conservation of Energy and a spring

    Wow, thanks for the very detailed explanation. I understand it now. Thanks again cepheid.
  18. M

    Conservation of Energy and a spring

    A 2.4kg block is dropped onto a spring with a force constant of 3.96*10^3 N/m from a height of 5m. When the block is momentarily at rest, the spring is compressed by 25cm. Find the speed of the block when the compression of the spring is 15cm. Relevant equations: External Work = change in...
  19. M

    Work and Kinetic Energy problem

    A force Fx acts on a particle that has a mass of 1.5kg. The force is related to the position x of the particle by the formula Fx = Cx^3, where C = .50 if x is in meters and Fx is in newtons. a) What are the SI units of C? b)Find the work done by this force as the particle moves from x=3.0m...
  20. M

    Collisions with rotations

    Hi, alphysicist, thanks for responding. The collision itself, the angular momentum is conserved. Angular momentum = I * omega Do I consider the before collision or after collision first? Do I use conservation of energy on the particle or the rod? Im just confuse on how to start this.
  21. M

    Collisions with rotations

    Anyone can help me get started with this problem please.
  22. M

    Collisions with rotations

    A uniform rod of length L1 = 2.2 m and mass M = 2.8 kg is supported by a hinge at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass m is supported by a thin string of length L2 = 1.8 m from the hinge. The particle sticks...
  23. M

    Rotation - Anuglar speed

    Thanks a lot Doc!
  24. M

    Rotation - Anuglar speed

    Oh, your right, Im solving for initial angular speed. So would Kf be 0? Since all we care if it gets barely over? Ui + Ki = Uf + kf MgR + 1/2(2MR^2)*omega initial ^2 = 2MgR + 0 Also Doc, when would I use the parallel axis theorem on problems? Like how would I know I should use it?
  25. M

    Rotation - Anuglar speed

    So Parallel Axis Theorem is I = Icm + Mh^2 I = MR^2 + MR^2 I = 2MR^2 So I can still apply the energy of conservation equation? Ui + Ki = Uf + Kf MgR + 0 = 0 + 1/2 (2MR^2)*omega ^2 MgR = 1/2 (2MR^2)*omega ^2 omega = square root of g/R this is the maximum angular velocity root of 9.81m/s/1.1m =...
  26. M

    Rotation - Anuglar speed

    Icm = MR^2 distance between the cm and parallel axis would be R right?
  27. M

    Rotation - Anuglar speed

    Oh, Doh! I was doing a same problem but with different numbers so please ignore the R = .75m, suppose to be R = 1.1m I went to tutoring and this is what the tutor did, I didn't really understood it. Parallel Axis Theorem is I = Icm + Mh^2 Im not sure what to do.
  28. M

    Rotation - Anuglar speed

    Rotation -- Anuglar speed A uniform ring 2.2 M in diameter is pivoted at one point on its perimeter so that it is free to rotate about a horizontal axis. Initially, the line joining the support point and center is horizontal. (a) If the ring is released from rest, what is its maximum angular...
  29. M

    Moment of Inertia and Torque

    Could someone check if what I did for part A/B is correct?
  30. M

    Moment of Inertia and Torque

    It says it gave the wheel an angular velocity of +610 rev/min to change this to rad/sec. I got 63.87905062 approx 64 rad/sec. 63.87905062 rad/sec / 24s = 2.661627109 approx 2.7 rad/s^2 is the acceleration Tnet = I alpha so Tnet/alpha = I 40 Nm /2.661627109 rad/s^2 = 15.02840110 approx 15 is the...
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