Recent content by MechTech

  1. M

    Projectile at 12 m height with v = 4 m/s at 42°: find position at 9 m

    You are forgetting you negative sign. A = -9.8
  2. M

    Projectile at 12 m height with v = 4 m/s at 42°: find position at 9 m

    From the problem we can assume that at t=0 the Y position is 12m. The last line ask us to solve from is 3/4 of Y or 0.75*12. Thus your equation for Y should look as follows: 0.75 *12 = 12 + 4* sin(42)*t - 9.8/2*t^2
  3. M

    Projectile at 12 m height with v = 4 m/s at 42°: find position at 9 m

    y_0 is 12 not zero. Y should be 3/4
  4. M

    What is the Ideal Current Source and How Does it Affect Circuit Design?

    Try redrawing the circuit with a straight center wire.
  5. M

    Projectile at 12 m height with v = 4 m/s at 42°: find position at 9 m

    Remember that the generic quadratic eq is ax^2+bx+c. You should get a positive number
  6. M

    Projectile at 12 m height with v = 4 m/s at 42°: find position at 9 m

    Think variable. Both functions are dependent on time.
  7. M

    What is the Ideal Current Source and How Does it Affect Circuit Design?

    Try redrawing the circuit
  8. M

    Projectile at 12 m height with v = 4 m/s at 42°: find position at 9 m

    You are on the right track with your attempt but you need to find a relationship between the two equations. (HINT: What do the x position and Y position equations have in common)
  9. M

    Need One More Class for Next Fall

    They are all elective courses. Thanks for your input
  10. M

    Need One More Class for Next Fall

    I am a rising senior Physics Major who needs one more class to meet the credit hour requirement to keep my scholarship. These are the class that fit into my schedule that my adviser told me to look at: Intro to Software Engineering - really just advanced C++, UML use, memory management, and...
  11. M

    Direction of sailboat velocity after wind gust

    try negative.33 and negative .3 since I would consider west negative x
  12. M

    Direction of sailboat velocity after wind gust

    Oh and here is a link to superpostion. It the pretty much they way you add vectors. http://en.wikipedia.org/wiki/Superposition_principle
  13. M

    Direction of sailboat velocity after wind gust

    You are on the right path. The cos(0)*.33=.33 that gives you the x component of V1. Sin(0)*.33 gives you the y component of V1. Now take the magnutiude of V2 (.3) and break it down into its X & Y component using -28 degrees (its negative because it south aka below the x axis). The add the X...
  14. M

    Direction of sailboat velocity after wind gust

    Use superpostion. And ad the vectors up. One vector has a magnutiude of .33m/s @ 0 degrees. The second is the wind blowing on the boat then use your arctan to find the angle
  15. M

    Solving for Force Q to Achieve Equilibrium for 2 Rods Connected by a Wire

    1. Collars A & B are connected by a 525-mm long wire and can slide freely on frictionless rods. If force P= (341 N)j is applied to collar A determine the tension of the wire when y=155mm and the magnitude of force Q(which is along the z axis) to maintain equilibrium 2. Homework Equations...
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