Recent content by cuj93

  1. C

    Propulsive force and force at interface

    So a) what is the propulsive force applied to the 6kg mass? 42N and b) what is the force at the interface between the two blocks? 6N
  2. C

    Propulsive force and force at interface

    Will that force be the force at the interface. Since 2 kg mass experiences no friction Fnet=ma Fnet=2(3) F=6N
  3. C

    Propulsive force and force at interface

    is this correct?
  4. C

    Propulsive force and force at interface

    Fnet=mneta F-18=mneta F-18=(6+2)(3) F-18=24 F=42N
  5. C

    Propulsive force and force at interface

  6. C

    Propulsive force and force at interface

    Homework Statement Two blocks, one of mass 6 kg and the other of mass 2 kg, are placed in contact on a horizontal table, and a constant horizontal force is applied to the 6 kg mass. There is a constant frictional force of 18 N between the 6 kg mass and the table, but no frictional force...
  7. C

    Sig Fig: Calculate (.82+.042)(4.4*10^3) - 3800

    Thank you everyone.
  8. C

    Sig Fig: Calculate (.82+.042)(4.4*10^3) - 3800

    That is what I am saying. The program says the correct answer is c.
  9. C

    Sig Fig: Calculate (.82+.042)(4.4*10^3) - 3800

    The program is not like a calculator. It gives the question then a list of answers and you select one. I can try to post a pic.
  10. C

    Sig Fig: Calculate (.82+.042)(4.4*10^3) - 3800

    Clue is in the number of significant figures they give you in the numbers provided. And they did a good job writing e.g. 4.4*103 instead of 44000 to make clear the number of significant digits... And you do get the correct idea writing .86 and not .862 ! So .82+.042=.86 according to sig...
  11. C

    Sig Fig: Calculate (.82+.042)(4.4*10^3) - 3800

    Homework Statement Calculate (.82+.042)(4.4*10^3) keeping only significant figures. The Attempt at a Solution I got .82+.042=.86 (.86)(4.4*10^3)=3784 So sig. fig says the answer should be 3800. Please explain what I did wrong. Thank you.
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