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  1. Q1 冇左figure 5.2. Q2 actually, to see the movement direction of particle in wave is very easy. when the wave travelling toward right, just see the point in the left hand side of this point. For Q, it is in 0.4cm RHS to the relative position, and in the next step, it will be same as...

  2. The answer should be A, only statement 2 is correct. The induced emf would make X at a higher potential than Y. Thus current flows from X to Y along the conducting wire (the external circuit) and from Y to X...

  3. Consider the horizontal component Horizontal speed of the ball =10cos30* =8.66 m/s Consider the vertical component Initial vertical speed of the ball =10sin30* =5 m/s The ball will experience a downward acceleration due to gravity. By the law of conservation of momentum...

  4. Reslove vertically: s=ut+a(t*2)/2 -500= -10(t*2)/2------------(u=0) t=10s Reslove horizontally: s=ut-----------------(a=0) d=200t d=2000m

  5. Buy a multi-socket adapter and open it. All metals are conductors. Where can find more information about "Circuits" go to http://www.vias.org/electronics.html

  6. (1)c---pq向右移...感生電流向上... 既然係1 o岩2就錯=.=.....3佢有emf先有current..所以都係o岩o既.. (2)b---t=4s時o既速度:2*4=8ms^-1 p=f*v p=900*8=7200w (3)b---撞擊中momentum改變(幅圖下o既面積):240*0.1/2=12kgms^-1 設撞擊後速度=xms^-1 (x+40)0.2=12 x=20ms^-1 (4)b---average force=momentum改變/時間 =12/0.1=120N

  7. By conservation of momentom, Initial momentom = final momentom Initial momentom= 2x2+ 1x-4 =0Ns Final momentom A: 2x-2+1X4=0Ns B:-1x2+1x2 =0Ns C;2x0+1x0=0Ns D:2x1+1x-2=0Ns E...

  8. 31) 這題可先計算物體落地時的速度及反彈速度。 落地的速度v1:(用機械能守恒公式計算) 0.5mv12 = mgh v12 = 2gh = 2x10x10 v1 = -14.14ms-1 (以向上為正) 反彈時的速度v2:(用機械能守恒公式計算) 0.5mv22 = mgh v22 = 2gh = 2x10x4 v2 = 8.94ms-1 Δv = v2 – v1 = 8.94 – (-14.14) = 23.08ms-1 (23.1ms-1...

  9. A shows the relationship of PE. and time when a object is fallen from a...

  10. 1.It shows the human reaction time. 2.It shows the reason why we need to prepare some...