close
標題:

Area & Volume

發問:

免費註冊體驗

 

此文章來自奇摩知識+如有不便請留言告知

A metal sphere of radius r is re-casted into n smaller spheres, find the percentage increase in surface area in terms of r and n.

最佳解答:

Thevolume of the original sphere = (4/3) π r3 The volume of each small sphere = (4/3) π r3 / n = (4/3) π (r / 3√n)3 The radius of each small sphere = r / 3√n The original surface area = 4 π r2 The new total surface area = n x [4 π (r / 3√n)2] = 4 π r2 3√n The percentage increase in surface area = [(4 π r2 3√n - 4 π r2) /4 π r2] x 100% = 3√n- 1 = 100(3√n - 1)% 2014-08-05 08:14:01 補充: The percentage increase in surface area is independent of r. 2014-08-05 08:45:49 補充: 兄台是英雄,小弟不是。 2014-08-06 00:27:27 補充: 你在說我嗎?怎麼我不知道。 小心!「聊天猜謎或其他非知識交流的行為」是檢舉的一個理由。哈哈!

其他解答:

重點:volume 不變。 A metal sphere of radius r has a volume of (4/3)πr3. That is re-casted into n smaller spheres, suppose the radius is s. Then, the total volume is n (4/3)πs3. That is, (4/3)πr3 = n (4/3)πs3, or r3 = ns3 2014-08-05 08:08:56 補充: The question wants to eliminate s, so we make s as the subject. s3 = r3/n s = ?(r3/n) = r/?n Old surface area = 4πr2 New surface area = n 4πs2 = n 4π (r/?n)2 = n 4πr2/n^(2/3) = 4πr2 n^(1/3) Thus, the percentage increase in surface area in terms of r and n is (New - Old)/(Old) × 100% 2014-08-05 08:11:32 補充: = [4πr2 n^(1/3) - 4πr2]/(4πr2) × 100% = [n^(1/3) - 1] × 100% (Ans) (Actually no need to be in terms of r. Only n is enough.) (Think of a special case. When n = 1, nothing is change, it makes sense.) 2014-08-05 08:13:51 補充: pingshek 兄,早晨~ 英雄所見略同! ﹝?????????﹞ 2014-08-05 18:18:11 補充: pingshek 兄,當日你幫助、遷就在下,再加入知識友,可見閣下的胸襟和風度。 閣下用心解答,回覆精準,對大家十分有幫助! 我是很感謝和欣賞你的! ☆ヾ(???)ノ 2014-08-06 16:35:11 補充: pingshek:對呀,是在說您。 謝謝提點,惡意檢舉的人的確很麻煩。 但我們也是在交流知識,這是禮儀文化的知識交流,正正是現今青少年最需要的。|||||唉我真係好想答架~但我竟然連f2的Maths都唔識做。好自卑! 2014-08-05 06:24:33 補充: 可否贊助你2點?我覺得答到嗰個好勁呀!
arrow
arrow

    香港美食推薦 發表在 痞客邦 留言(0) 人氣()