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物理 高中

想請問第二小題物理,謝謝大家🙏

Yo <t. 由平面運動斜拋理論所計算的答案與實際數據不相符的原因,你認為可能有哪些? (服加速度以9.8 m/s² 計,1 m = 3.28 ft) 022-3个 空氣阻力etc 2. 電腦模擬的運動軌跡 在現代的科學研究中,實驗、理論解析、電腦模擬相輔相成,讓我 們得以瞭解更多自然的現象與規律。因此除了解析推導外,也可以 運用簡單的程式,來計算物體運動的軌跡。由於物體受力大多僅與 其位置有關,我們可以藉由以下的運算法,來計算物體運動的軌 跡: y(m) 1.0F 0.8 0.6 23 0.4 0.2 10-60°° 由物體的初始位置,已知所受外力,進而求出該時刻的加速度。 ②利用求得的加速度,可由物體的初始速度,算出下一瞬間的速度。 3 藉由算出的速度,可由起始位置,算出下一瞬間的位置。 ④重複以上的計算步驟,即可求出物體運動的軌跡。 撰寫電腦程式計算時,我們無法取無限小的時距,因此嚴格來說,數值計算得到的是近似的軌跡 點。但只要時距取得夠短,這些軌跡點就會趨近正確的解答。如上圖所示,物體以相同初始速率作斜 拋運動,當拋射角度為 30° 與 60° 時,雖然運動的軌跡不同,但其水平射程相同。此結果與解析推導 相同,可以互為佐證,但許多實際的問題並沒有解析解,若能善用電腦程式計算,就能提供另一種理 解的管道,是非常強大有用的工具。 【混合題】 (1) 分析拋射角度為 60° 的軌跡時,會注意到軌跡點在頂點處較密,而在起始位置或落地處較疏。但若 分析拋射角度為30° 的軌跡時,其軌跡點則較為均勻。針對軌跡點的疏密變化,以下哪些為合理 的推論?(多選) (A)物體在頂點處的速率最小(B)物體在起始位置或落地處的水平速度較小 (C)物體在頂點處的水平速度較小(D)拋出至最高點期間,拋射角度大時,物體的速率變化較大,故 軌跡點疏密變化也較大(E)抛射角度小時,重力加速度較小,故軌跡點較為均勻。 0=30° 0 10.5 1.0 1.5 2.0 (m) (2) 進行電腦程式計算時,每一軌跡點的時距皆相等。若拋射角度為 30°時,其飛行時間為 10.0 s,試 由分析上圖中的軌跡點,估算拋射角度為 60°時的飛行時間為何? 17.55 +ZON 88 STATIONERY CORP LIBERTY

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自然科學 大學

想問3-9的a跟c

Problems 3-1. Explain the difference between *(a) random and systematic error. (b) constant and proportional error. *(c) absolute and relative error. (d) mean and median. *3-2. Suggest two sources of systematic error and two sources of random error in measuring the length of a 3-m table with a 1-m metal rule. 3-3. Name three types of systematic errors. *3-4. Describe at least three systematic errors that might occur while weighing a solid on an analytical balance. *3-5. Describe at least three ways in which a systematic error might occur while using a pipet to transfer a known volume of liquid. 3-6. Describe how systematic method errors may be detected. *3-7. What kind of systematic errors are detected by varying the sample size? 3-8. A method of analysis yields masses of gold that are low by 0.4 mg. Calculate the percent relative error caused by this result if the mass of gold in the sample is (a) 500 mg. (b) 250 mg. V(c) 125 mg. (d) 60 mg. 3-9. The method described in Problem 3-8 is to be used for the analysis of ores that assay about 1.2% gold. What minimum sample mass should be taken if the relative érror resulting from a 0.4-mg loss is not to exceed *(a) -0.1%? (b) -0.4%? (c) -0.8%? (d) - 1.1%? 3-10. The color change of a chemical indicator requires an overtitration of 0.03 mL. Calculate the error if the total volume of titrant is percent relative (a) 50.00 mL. (c) 25.0 mL. 3-11. A loss of 0.4 mg of Zn occurs in the course of an percent relative analysis for that element. Calculate the error due to this loss if the mass of Zn in the sample is *(b) 10.0 mL. (d) 30.0 mL. 190 (c) 188 (d) 4.52 x 103 4.63 x 103 4.53 x 10 ³ √6 *(a) 30 mg. (b) 100 mg. *(c) 300 mg. (d) 500 mg. 3-12. Find the mean and median of each of the following sets of data. Determine the deviation from the mean for each data point within the sets, and find the mean devi- Vation for each set. Use a spreadsheet if it is convenient. *(a) 0.0110 0.0105 (b) 24.53 0.0104 24.68 24.81 24.77 39.61 862 (f) 850 MA 3-13. Challenge Problem: Richards and W the molar mass of lithium and colle data. 24.73 Experiment 1 2 3 4 5 6 194 447 X 10 7 448 X 107 4.58 X 10 (a) Find the mean molar t workers. (b) Find the median molar ma (c) Assuming that the cam molar mass of lithium is the absolute ertor and of the mean value demi Willard. (d) Find in the chemical ues for the molar mus since 1910, and ag a table or spreadshera 1817 given in the a Richards and Willd. Com mass versus year to la of lithium has chang Suggest possible abruptly about 18 ant de (e) The incredibly deals Richards and W that major changes will occur. Disc calculation in pat (f) What factors ha since 1910? (g) How would you mass? 6See Chapter 2 of Applications of Microsoft Excel in Analytical Chemistry, 4th ed., for information about statistical 7T. W. Richards and H. H. Willard, J. Am. Chem. Soc., 1910, 32, 4, DOI: 10.1021/ja01919a002. built-in statistical functions. "Answers are provided at the end of the book for questions and problems marked with an asterisk The I of ₂ or inc rce of able vas error c often in individu ate resul data in rtainties. dimensio andom e analysts The result

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