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  1. (1) Area of gate = 2 x 1.5 m^2 = 3 m^2 Depth of centroid of gate from water surface = (3 + (1.5/2)sin(25)) m = 3.317 m Pressure at centroid = 1000 x 9.81 ...

  2. Angle between AB and the horizontal = arc-tan(1.5/2) degrees = 36.87 degrees Lenth of AB = square-root[1.5^2 + 2^2] m = 2.5 m The centroid of AB is at its ...

  3. Distance of centroid of semi-circle from bounding diameter = 2 x 2/3.pi m = 0.4244 m Distance of upper semi-circle centroid from top of tank = (1 - 0.4244) ...

  4. Force due to water acting on the gate = 1000 x 9.81 x (h/2) x (1.25 x h/sin(75)) = 6348h^2 where 1000 kg/m^2 is the density of water 9.81 m/s^2 is the ...

  5. Dept of A below surafce of water at the left hand side of the gate = (4 - 2.sin(60)) m = 2.268 m Upward thrust acting on the gate by the water to its left ...

  6. Fan 吹出來的風會使到 trolley 受到一個向右的力。但當風吹在 card 上又使 trolley 受到一個向左的力,如果理想的話,两個力會相消,trolley 唔郁,現實可能向右方郁 ...

  7. (a) Change of momentum of the ball = (140/1000) x (-39 - 39) kg.m/s = -10.92 kg.m/s Because impulse = change of momentum, thus magnitude of impulse = 10.92 ...

  8. Let L be the length of the rope. Use conservation of energy, (1/2)mv^2 = mgL where v is the speed of the bob at the lowest position g is the acceleration ...

  9. ...igloo surface = R - R.cos(a) = R(1 - cos(a)) By conservation of mechanical energy, (1/2)mv^2 = mg.R(1-cos(a)) v^2 = 2gR(1 - cos(a)) Thus...

  10. Work done by gravitational force = 3g x (10 - 2) J = 240 J Distance travelled by the stone along the plane = (10 - 2)/sin(30) m = 16 m Frictional force = ...