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匿名 發問於 科學及數學其他 - 科學 · 7 年前

S.4 Phy

1. A monkey falls from rest from a tall tree. How far does it fall during the third second? Assume air resistance is negligible. Answer is 25m. Why?

2. A car keeps slowing down uniformlyalong a straight road. It slows down to 20 ms^(-1) after travelling a distance of 100 m, and runs 80 m further before it comes to a stop. What is its speed at first? Answer is 30 ms^(-1). Why?

3. CN Tower in Canada is 533 m tall and it is the tallest building in the World. If a stone of 200 g is thrown vertically upwards at 5 ms^(-1) at the top of the CN Tower, what is the kinetic energy of the stone when it drops to the groung? Neglect air resistance. Answer is 1069J. Why?

更新:

我睇唔明Q1 and 2

:(

1 個解答

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  • CK
    Lv 7
    7 年前
    最愛解答

    3.   When the small stone was thrown. it had a K.E. of (1/2)(0.2)(5²). It goes up and change this K.E. to P.E.

        When the small stone comes down to a height of 533 m. the P.E. gained returns back to K.E.

        The total energy of the small stone is (1/2)(0.2)(5²) + mgh = (1/2)(0.2)(5²) + (0.2)(10)(533) = 1068.5 J

    2014-03-09 17:05:37 補充:

    1.    A monkey falls from rest from a tall tree. How far does it fall during the third second? Assume air resistance is negligible. Answer is 25m. Why?

        In the first 2 seconds, It falls : h = V₀t + (1/2)(g)t² = (0)(2) + (1/2)(10)(2)² = 20 m

    2014-03-09 17:07:02 補充:

        In the first 3 seconds, It falls : h = V₀t + (1/2)(g)t² = (0)(3) + (1/2)(10)(3)² = 45 m

        It falls (45 - 20) = 25 m in the third second

    2014-03-09 17:11:08 補充:

    2.    A car keeps slowing down uniformlyalong a straight road. It slows down to 20 ms^(-1) after travelling a distance of 100 m, and runs 80 m further before it comes to a stop. What is its speed at first? Answer is 30 ms^(-1). Why?

        Use (Vf)² = (Vi)² - 2aS

        20² = (Vi)² - 2a(100) ... (1)

    2014-03-09 17:15:25 補充:

         It comes to a stop means V=0

         Sub into the formula again

         0² = 20² - 2a(80)

         a = 2.5 m/s²

         put (a = 2.5 m/s²) into (1) ==>

         20² = (Vi)² - 2(2.5)(100)

         Vi = 30 m/s

    2014-03-13 20:30:01 補充:

    1.  猴子從高樹跌下,在第 3 秒下跌了多小米。

       0 至 1 秒是 第 1 秒

       1 至 2 秒是 第 2 秒

       2 至 3 秒是 第 3 秒

       在頭 2 秒 下跌了 h = V₀t + (1/2)(g)t² = (0)(2) + (1/2)(10)(2)² = 20 m

       在頭 3 秒 下跌了 h = V₀t + (1/2)(g)t² = (0)(3) + (1/2)(10)(3)² = 45 m

       在 第 3 秒 下跌了 (45 - 20) = 25 m

    2014-03-13 20:36:24 補充:

    2.  小車在直路上持續减速,它在行駛 100 米後減速至 20 米/秒,再前進 30 米之後完全停定,求小車初速。

    公式 : (Vf)² = (Vi)² - 2aS

    其中 :

    Vf = 末速

    Vi = 初速

    a = 加速度

    S = 位移 (前進距離)

    2014-03-13 20:39:19 補充:

    用 行駛 100 米後減速至 20 米/秒 代入

    20² = (Vi)² - 2a(100) ... (1)

    用 再前進 80 米之後完全停定 代入

    停定 即 Vf = 0

         0² = 20² - 2a(80)

         a = 2.5 m/s²

         用 (a = 2.5 m/s²) 代入 (1) ==>

         20² = (Vi)² - 2(2.5)(100)

         Vi = 30 m/s

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