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a1 solution - Question1 Question2(a...

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Question 1 Question 2 (a) volume (in litres) = depth (in m) * 4 m * 5 m * 1000 (litres / m 3 ) depth (in m)= volume (in litres)/ 20000 pressure at bottom (in Pa) = depth(in m) * density of water (in kg /m 3 ) * g (in m/s 2 ) pressure (in Pa) = (volume / 20000) * 998 * 9.81 = volume (in litres) * 0.489 Pa/litre 0 0007335 . 0 0 489 . 0 ) 489 . 0 * ( V V V k V V k V o o Note that the 0.0007335 in the above has units of 1/seconds.
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The system is analogous to the rotating mass and discharging capacitor examples. / 0 V V(t) seconds 1363 0007335 . 0 / 1 t e 6 minutes = 360 seconds litres 1152 1500 V(360) 1363 / 360 e To find time to 100 litres, set V ( t ) equal to 100 seconds 691 , 3 ) 1500 / 100 ln( 1363 ) 1500 / 100 ln( 1363 / 1500 / 100 litres 100 1500 V(t) 1363 / 1363 / t t e e t t (b) The defining equation becomes R V V 0007335 . 0 Note that all of the terms in the equation have units of litres/second. The system is now analogous to the charging capacitor example. ) -e kR( V(t) k k t/ τ 1 seconds 1363 ) units no ( 1 / seconds 1363 0007335 . 0 / 1 Plugging t = 100 seconds and R
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a1 solution - Question1 Question2(a...

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