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  1. ...the vertical motion of the ball after it is released, use: s = ut + (1/2) at ^2 with s = 0.5 m, u = 0 m/s, a = -g(=-10 m/s^2), t=? 0.5 = (1/2).(10)t^2 t...

  2. ... space, an isolated -ve charge produces -ve potential at its surrounding space. Thus, a +ve test charge under a +ve potential...

  3. 1. Apply the equation: s = ut + (1/2) at ^2 to the vertical motion from R to S with s = -10m... is wrong because when the 2nd order max is at 15 degrees, the max order of fringe is 7, not 4...

  4. ...3)^(2/3) AU = 17.83 AU (d) Because the distance from the sun at Q is shorter than that at R, the pontential energy of comet...

  5. ...).l = NBl^2J (c) Torque produced by the object at the axle of the motor = mg x (6/100) = 0.06mg = 0.6m where g is the acceleration...

  6. ..., (c/co)^2 = 750/300 c/co = square-root(750/300) = 1.58 (c) New pressure at partition A, Pa = (n1)R(300)/(L1)A New pressure...

  7. ...1.36 hence, c = 47.33 degrees (b) Let the refracted angle at the other side (inside the soap) at A be r r = (c - 10) degrees = (47...

  8. ...of the needle is pointing towards the back for the compass placed at the centre of the coil. For the other two, just treat the (small) section...

  9. ...know that the two fragments are moving at opposite direction. (Each fragment has linear momentum...the answer made by Gabriella Montez ~ mock exam in progress" ^^ 希望你能明白我的邏輯,要完全...

  10. ...distance between the highest point reached and the point reached at t1 be s by v^2-u^2=2as 0-(3)^2 = 2(-10)(s) s = 0.45m But the distnace...