Recent content by marialo

  1. M

    Adiabatic heat exchanger problem

    Adiabatic heat exchanger problem... Please Help! I have this proplem as a homework assignment and I'm a bit stuck as houw to set it up. Here it is: Propane gas enters a continuous adiabatic heat exchanger at 40 degrees C and 250 kPa and exits at 240 degrees C. Superheated steam at 300...
  2. M

    Solve F for 400N Post on 36.9° Rope

    Ok, i figured it out i think, please check me: The sum of the torques taken at the pivot point is equal to n(h)-Tcos(36.9)=0, which gave me a tension of 500 N. I then used this to find that the force is equal to 420 N. But what if the force is applied 6/10 of the way from the ground to the...
  3. M

    Solve F for 400N Post on 36.9° Rope

    I can't seem to figure out this question: One end of a post weighing 400 N and with height h rests on a rough rope fastened to the surface and making an angle of 36.9 degrees with the post. A horizontal force is excerted on the post. If the force (F) is applied at the midpoint of the post...
  4. M

    Calculating Coefficient of Friction and Kinetic Energy in Bullet-Block Collision

    Sorry, I'm still having trouble seeing how to go from KE to mu
  5. M

    Calculating Coefficient of Friction and Kinetic Energy in Bullet-Block Collision

    The work done be friction is d*mg(mu), where mg(mu) is the force of friction. the block's KE is just 1/2mv^2, right?
  6. M

    Canoe Movement Calculations: 45kg Woman in a 60kg Canoe

    i'm having a problem with this one: 45kg woman is in a 60kg canoe of length 5m. They are initially at rest. she walks from a point 1m from the end to a point 1m from the other end where she stops. If resistance and the motion of the canoe in the water are ignored, how far does the canoe move...
  7. M

    Calculating Coefficient of Friction and Kinetic Energy in Bullet-Block Collision

    i already did that. since momentum is conserved, the momentum of the bullet initially is equal to the sum of the momentums after the collision. i found that the velocity off the block after collision was 1.4 m/s.
  8. M

    Calculating Coefficient of Friction and Kinetic Energy in Bullet-Block Collision

    Sorry, i forgot to metion that the bullet is 4.0 g, or 0.004 kg
  9. M

    Calculating Coefficient of Friction and Kinetic Energy in Bullet-Block Collision

    I'm having trouble with this physics problem: A bullet traveling horizontally with a velocity of magnitude 400 m/s, is fired into wooden block with mass 0.800 kg initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 120 m/s. the...
  10. M

    Solving 3 Sphere Collision Problem: Help Needed

    I'm having trouble with this problem, can someone help: Spheres A (m=0.02 kg), B (m=0.03 kg), and C (m=0.05 kg), are each approaching the origin as they slide on a frictionless air table. the initial velocity of A is 1.5 m/s horizontally to the left and the x and y components of B's velocity...
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