我想問關於variations,唔該你地幫下我啦...

2008-04-04 9:46 pm
1.It is given that V varies jointly as r 2次 and h
SUB V=108 r=3 and h=4
a) Find an equation connecting h,r,V.
b) If r is halved and h is doubled, find the ratio of the new value of V to its
orginal value.
我識計a,但唔識b

2.The time (x min) required to boil a kettle of water varies directly as the
volume(V cm3)of water and inversely as the power(P W) of the stove.
It is given that the volumes of two kettles of water re in the ratio 9:13
and the power of the stove used to boil them respectively are in the ratio
12:13.If 9 minutes is required to boil the first kettle of water,find the time
required to boil the second kettle.

回答 (1)

2008-04-04 11:04 pm
✔ 最佳答案
1.(a)
let k be the variation constant
V=khr^2
108=k*4*3^2
108=36k
k=3
i.e.V=3hr^2

(b)
let R,H,v be the new r,h,V respectively
R=r/2
H=2h

v=3*2h*(r/2)^2
v=(3hr^2)/2

v:V
=(3hr^2)/2 : 3hr^2
=(3hr^2)/2 * 1/(3hr^2)
=1:2



2.
let k be the variation constant
x=kV/P

代入第一壺水d數據

9=9k/12
9=3k/4
k=12

求得k
再代入條公式:

x=12V/P

再代返第二個壺d數據入去,就求到第二壺水既boil time啦

boil time required for the 2nd kettle = 12*13/13
=12mins

2008-04-04 15:14:57 補充:
第二題,簡單起見可以設2個壺既個volume同power分別係9單位,13單位 & 12單位,13單位
都做到

不過如果你要用黎交功課就要用返formal d既做法
設佢為9a單位,13a單位 & 12a單位,13a單位

因為佢只話個比例係咁,冇比個實際數據你
e.g.可以係18單位,26單位 & 24單位, 26單位

2008-04-05 17:22:36 補充:
其實冇咩野,只係我呢d懶人既偷雞做法,雖然佢只比左個比(冇單位),但你可以設實際數據就係咁,因為只要符合佢個比,計出來的答案都會係果一個,唔會變,但你考試時要用正規的寫法(否則會扣pp),做MC時先用,考試就寫let the volume & power of the 1st kettle be 9a(units) & 12a(units)...再代入得到9=9ak/12a,而由於分子分母約左,答案同上面一樣.

2008-04-05 17:22:41 補充:
另外唔一定要用9a,12a,你鍾意既用9b,12b都得(個字母代表變數),因為無論代入去咩數,佢地既比值都唔會變,計出來的都會係符合題意的1pair,呢個就係所謂正規的做法(我上面就係簡單d直接代入個1)


收錄日期: 2021-04-14 20:49:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080404000051KK01224

檢視 Wayback Machine 備份