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Posted by on Wednesday, October 24, 2012 at 6:17pm.

Forces of 11.1N north, 19.9N east, and 14.0N south are simultaneously applied to a 3.62kg mass as it rests on an air table. What is the magnitude of its acceleration?

b) What is the direction of the acceleration in degrees? (Take east to be 0 degrees and counterclockwise to be positive. Enter an angle between -180 degrees and +180 degrees.)

  • Physics HELP!! - , Thursday, October 25, 2012 at 2:41am

    F(x) =F1(x)+F2(x)+F3(x) =0+19.9 +0 =19.9 N
    F(y) =F1(y)+F2(y)+F3(y) =11.1+0-14=-3 N
    F=sqrt{F(x)^2 + F(y)^2}
    a=F/m
    tan alpha=F(y)/F(x)
    unknown angle =appha +90 degrees

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