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Question-68510




Question Number 68510 by oyemi kemewari last updated on 12/Sep/19
Answered by mr W last updated on 14/Sep/19
v=(√(2gh))  base area of tanks=A=1 ft^2   A(dh/2)=−((πd^2 )/4)vdt  dh=−((πd^2 (√(2gh)))/(2A))dt  (dh/( (√h)))=−((πd^2 (√(2g)))/(2A))dt  ∫_h_0  ^h (dh/( (√h)))=−((πd^2 (√(2g)))/(2A))∫_0 ^t dt  2[(√h)−(√h_0 )]=−((πd^2 (√(2g)))/(2A))t  (√h_0 )−(√h)=((πd^2 (√(2g)))/(4A))t  ⇒t=((4A((√h_0 )−(√h)))/(πd^2 (√(2g))))  h_0 =1 ft  d=0.5 in=(1/(24)) ft    for h=0.25 ft:  t=((4×1((√1)−(√(0.25))))/(π((1/(24)))^2 (√(2g))))=82 seconds    for h=0:  t=((4×1((√1)−(√0)))/(π((1/(24)))^2 (√(2g))))=164 seconds
v=2ghbaseareaoftanks=A=1ft2Adh2=πd24vdtdh=πd22gh2Adtdhh=πd22g2Adth0hdhh=πd22g2A0tdt2[hh0]=πd22g2Ath0h=πd22g4Att=4A(h0h)πd22gh0=1ftd=0.5in=124ftforh=0.25ft:t=4×1(10.25)π(124)22g=82secondsforh=0:t=4×1(10)π(124)22g=164seconds
Commented by mr W last updated on 14/Sep/19
Commented by oyemi kemewari last updated on 17/Sep/19
Commented by oyemi kemewari last updated on 28/Sep/19
thank you sir

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