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第一章 量 測

第一章 量 測. 1-1 物理學是什麼? 1-2 測量 1-3 國際單位系統 1-4 單位換算 1-5 長度 1-6 時間 1-7 質量. 1-1 物理學是什麼?. 從 『 物理 』 、 『 格物 』 到 〝Physics〞 『 細推物理須行樂,何用浮名絆此生。 』- 李政道 海闊天空皆 『 物理 』 物理學是 『 活 』 的科學。. 1-2 測 量. 物理量:大小 + 單位. 測量:物理學的基礎. 需定出一“標準”,以便參考比較. 標準訂定的要求:. 1. 易於取得。 2. 恒定不變。. 1-3 國際單位系統.

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第一章 量 測

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  1. 第一章 量 測 1-1 物理學是什麼? 1-2 測量 1-3 國際單位系統 1-4 單位換算 1-5 長度 1-6 時間 1-7質量

  2. 1-1 物理學是什麼? • 從『物理』、 『格物』到〝Physics〞 • 『細推物理須行樂,何用浮名絆此生。 』-李政道 • 海闊天空皆『物理』 • 物理學是『活』的科學。

  3. 1-2 測 量 物理量:大小+單位 測量:物理學的基礎 需定出一“標準”,以便參考比較 標準訂定的要求: 1.易於取得。 2.恒定不變。

  4. 1-3國際單位系統 基本單位:共七種。 國際單位(SI單位): 1971年第14屆國際度量衡大會所制定,又稱公制系統 。 分基本單位與導出單位兩大類。 導出單位:由基本單位定義出之物理量。

  5. 表1.2 SI 單位的簡寫

  6. 換算因子 1-4單位換算 鏈鎖變化法:將原始數據乘一換算因子,以達到單位換算之目的。 Exp:將2分鐘換成以秒為單位。

  7. 光年:光行進一年所花的時間 1 光 年 = 幾公尺?

  8. 1-5 長 度 1792年:米是北極到赤道距離的千萬分之一 1959年:碼是米的0.9144倍 1960年:米是氪85原子所發射的橘紅色光波 長的1,650,763.73 倍 1983年:米是光在真空中進行299,992,458分 之一秒所行的距離

  9. 電子顯微鏡下的病毒照片 由病毒細胞攫取的脂蛋白(淺色部分)環繞著病毒的核心(深色部分)。 全部直徑小於50nm。

  10. 1-5 時 間 1.一秒為地球自轉週期的315576000分之一 2.一秒為銫133原子所發出某依特定波長的光波頻率的 9,192,631,770倍

  11. 幽靈般的銀藍色雲朵

  12. 到太陽頂端的視線 第一次日落 遠方的太陽 第二次日落 地球中心 利用日落計算幽靈雲的高度 由(θ/360o)= (t/24h) t = 38min代入,得θ=9.5o 由圖可知 cosθ= r/(r+h) ∵ r = 6.37×106m,θ= 9.5o 代入即可得 h = 8.86×104m~89km 若日落後38min仍看得到幽靈雲,但很快就變黯淡了,則其高度h約為若干?

  13. 1-7 質 量 1 u = 1.6605402 × 10-27 kg 1. 一仟克(kg)由依各保存在巴黎附近的鉑銥原型所訂定。 2.一原子質量單位(u)為碳12原子質量的12分之一

  14. 習:7, 16, 23

  15. 7. The volume of ice is given by the product of the semicircular surface area and the thickness. The are of the semicircle is A = pr2/2, where r is the radius. Therefore, the volume is where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have In these units, the thickness becomes which yields,

  16. 16. We denote the pulsar rotation rate f (for frequency). (a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figure considerations for a moment), we obtain the number of rotations: which should now be rounded to 3.88×108 rotations since the time-interval was specified in the problem to three significant figures.

  17. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or which yields the result t = 1557.80644887275 s (though students who do this calculation on their calculator might not obtain those last several digits). (c) Careful reading of the problem shows that the time-uncertainty per revolution is . We therefore expect that as a result of one million revolutions, the uncertainty should be .

  18. 23. We introduce the notion of density, r=m/V , and convert to SI units: 1000 g = 1 kg, and 100 cm = 1 m. (a) The density r of a sample of iron is therefore which yields r = 7870 kg/m3. If we ignore the empty spaces between the close-packed spheres, then the density of an individual iron atom will be the same as the density of any iron sample. That is, if M is the mass and V is the volume of an atom, then

  19. (b) We set V = 4pR3/3, where R is the radius of an atom . Solving for R, we find The center-to-center distance between atoms is twice the radius, or 2.82 × 10-10 m.

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