Ｐ=mgh(h-x)
h做返主項
要有步驟
thx!!!
一條中二既數...好多中6-7既人都唔識(到e家仲未解)10分!!
2007-01-05 4:31 am
回答 (9)
2007-01-05 4:43 am
✔ 最佳答案
Ｐ= mgh ( h - x )
h - x = mgh / P
h - x = mg/P ( h )
( h - x ) / h = mg/P
h/h - x/h = mg/P
1 - x/h = mg/P
x/h = 1 - mg/P
x/h = ( P - mg ) / P
h ( P - mg ) = xP ... 交义相乘
h = xP / ( P - mg )
2007-05-03 7:30 am
這個最佳回答第一步已經錯了！請留意一下：
P=mgh(h-x)下一步頂多是:
1/(h-x)=mgh/P
不過這樣都是計不到的
但至少要知道這個答案不是對的!!!!!!
P=mgh(h-x)下一步頂多是:
1/(h-x)=mgh/P
不過這樣都是計不到的
但至少要知道這個答案不是對的!!!!!!
2007-01-26 11:16 pm
the answer by y99shto is correct
2007-01-12 9:43 pm
Ｐ= mgh ( h - x )
h - x = mgh / P ??????????
(你計錯數了，應該是︰h-x = P / mgh)
h - x = mgh / P ??????????
(你計錯數了，應該是︰h-x = P / mgh)
2007-01-12 6:38 pm
Ｐ=mgh(h-x)
P/h-x=mgh
P=(mgh)(h-x)
P=h(mg‧-x)
h=P/-mgx
P/h-x=mgh
P=(mgh)(h-x)
P=h(mg‧-x)
h=P/-mgx
2007-01-05 5:59 am
I'll use a method called "completing square". This method does not require quadratic formula. Thus a form two student can easily understand.
h(h-x)=P/mg
h^2 - 2(x/2)h + (x/2)^2 = P/mg + (x/2)^2 <---Add (x/2)^2 on both sides
(h - x/2)^2 = (4P+mg)/4mg
Square rooting both sides,
h-x/2 = 士√{(4P+mg)/4mg}
h = x/2 士 √{(4P+mg)/4mg}
Thus the answer is h = x/2 士 √{(4P+mg)/4mg}
h(h-x)=P/mg
h^2 - 2(x/2)h + (x/2)^2 = P/mg + (x/2)^2 <---Add (x/2)^2 on both sides
(h - x/2)^2 = (4P+mg)/4mg
Square rooting both sides,
h-x/2 = 士√{(4P+mg)/4mg}
h = x/2 士 √{(4P+mg)/4mg}
Thus the answer is h = x/2 士 √{(4P+mg)/4mg}
2007-01-05 4:50 am
這是一條二次方程 (quadratic equation)，用配的方法可解。
P =mgh(h-x)
mg(h² - hx) = P
mg[h² - 2h(x/2) + (x/2)² - (x/2)²] = P
mg(h - x/2)² - mg(x/2)² = P
mg(h - x/2)² - mgx²/4 = P
mg(h - x/2)² = P + mgx²/4
mg(h - x/2)² = (4P + mgx²)/4
(h - x/2)² = (4P + mgx²)/4mg
h - x/2 = ±√[(4P + mgx²)/4mg]
h = x/2 ± √[(4P + mgx²)/4mg]
h = x/2 + √[(4P + mgx²)/4mg] 或 h = x/2 - √[(4P + mgx²)/4mg]
P =mgh(h-x)
mg(h² - hx) = P
mg[h² - 2h(x/2) + (x/2)² - (x/2)²] = P
mg(h - x/2)² - mg(x/2)² = P
mg(h - x/2)² - mgx²/4 = P
mg(h - x/2)² = P + mgx²/4
mg(h - x/2)² = (4P + mgx²)/4
(h - x/2)² = (4P + mgx²)/4mg
h - x/2 = ±√[(4P + mgx²)/4mg]
h = x/2 ± √[(4P + mgx²)/4mg]
h = x/2 + √[(4P + mgx²)/4mg] 或 h = x/2 - √[(4P + mgx²)/4mg]
2007-01-05 4:45 am
P/(h-x)=mgh
p/h-p/x=mgh
p-ph/x=mghh
px/x-ph/x=mghh
p(x-h)/x=mghh
x-h=mghhx/p
.....
唔識
p/h-p/x=mgh
p-ph/x=mghh
px/x-ph/x=mghh
p(x-h)/x=mghh
x-h=mghhx/p
.....
唔識
參考: me
2007-01-05 4:35 am
係咪物理來架?
mgh
係先有得計
唔係就真係唔識
mgh
係先有得計
唔係就真係唔識
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