a swiming pool is bein filld at a rate of 1 liter per second,how long wil it take 2 fil a 6000 liter pool_min?
463155
6000 seconds\n6000 seconds = 6000/60 minutes = 100 minutes\n100 minutes = 1 hour, 40 minutes
687277
1 hour or 60 mins or 3600 seconds :-) is the time it takes but it will still be 12:30 :-p
748998
Assume smaller pipe fills the tank in x hours and the larger one in (x-2.5) hours. Taking the given inf that both together can fill the tank in 8.4 hours, develop a quadratic equation in x, and solve it. \n\nLarger pipe fills the tank in 15.65 hours whereas the smaller one takes 18.15 hours to fill it.
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
953838
1 second = 1 000 000 microseconds
972014
to solve the given problem, all you need to apply is the conservation of momentum\nthe formula is m1u1+m2u2=m1v1+m2v2\nm1,m2 are masses of the bodies, in this case, the fish\nu1,u2 are initial velocities of the bodies\nv1,v2 are final velocities of the bodies\nthe masses of the fishes are m1=5Kg, m2=1Kg\n\n1) here, initial velocity of the small fish is zero, so u2=0. u1=1m/s\nsince the larger fish swallows the smaller fish, after collision, they coalesce to form a single fish. so, the equation would be\nm1u1+m2u2=(m1+m2)v\n5*1+1*0=(5+1)v\n5=6v => v=5/6m/s\n\n2) by convention, we take the velocity of the body moving towards right to be positive. so we take the velocity of the larger fish to be positive. so, the velocity of the smaller fish would be negative.\napplying the law of conservation of momentum,\nm1u1-m2u2=(m1+m2)v\n5*1-1*4=(5+1)v\n1=6v => v=1/6m/s\n\n3) here, the smaller fish is swimming twice as fast, ie., with a velocity of 8m/s towards the left\nso, m1u1-m2u2=(m1+m2)v\n5*1-1*8=(5+1)v\n-3=6v => v= -1/2m/s\nthis means that the larger fish, after swallowing the smaller fish, will be pushed back with a velocity of 0.5m/s.
983684
Well on the varsity swim team, in a two hour practice we swim round 6000-7000 yards in 2 hours.\nSets will vary, but usually we will do a long set (consisting of 1000 yards, etc. 10x100 on let's say...1:30. \nThen we'll do a shorter set, say 8x75's on 1:30.\nThen a long one, 3x500 on...7 min.\nThen @ the end of practice (last 15 minutes or so) we might play a fierce game of water polo!!!!!!!!!!!!
901904
I think here is a mistake:\n\n(1500 * 36) = 1000v\nv = 55.8 m/s\n\nyou have to put 500 instead 1500, because this is what we have after 1000 left us!\n\n(500*36)=1000v\nv=15 m/s
958697
10.4 m/sec\n\nV(t) = (30+a*t) m/sec\n\nwhere a = - 9.8m/sec^2\n\nV(2) = (30 - 09.8*2) m/sec\n\nV(2) = 30-19.6 = 10.4 m/sec
708766
Say that a boy runs 1000 km in 5 minutes.  A unit rate must have the second quantity at 1 unit, so divide both by 5 to get 200 km per min.\n\nThe height decreases so it shows a negative rate of change, perhaps -2 mm  per minute
131602
It's only a mile for crying out loud.  Don't worry... if you can swim 2000 yards in a regular pool without stopping at a pace of around 50 seconds per 50 you will have no problem.
758481
It's 8 meters per second.  He's gone 5000 meters in 625 seconds; 5000 meters/625 seconds = 8 meters per second.\n\n v=d/t\nv=5000 m/625 seconds\nv=8m/sec
847968
About 2 litres is about right as a minimum...;
581128
In order to achieve 600 ppm, \n600/120 pound= 5 pound product should be added to 1000 gal water.\n∴ In 10 gallon water, 5 pound * 10/1000 = 0.05 pound of product should be added. This is to be added in 30 seconds.\n\nHence, the flow rate of product is: 0.05/(1/2) = 0.1 lb/minute = 0.1 *60*24 lb/day = 144 lb/day * 1 gal/10 lb = 14.4 gal/day \n  \n\n⇒ %output = (14.4 gal/day)/(22 gal / day) = 65.45.. ←Answer
611115
Do you mean ML/hr (mega litres per hour) OR nL/hr (nano litres per hour)?\n\n1 litre  = 0.035315 cubic feet and of course 60 min = 1 hour\n\nSo 1 ML/hr = 0.035315 x 10^6 / 60 = 588.6 cubic feet per min
139303
the physics formula that can be applied here is:\n\nvt = x, where v = speed tub is being filled, t = minutes, x = number of tubs filled.\n\nsince there are three valves, each contributes to the filling/emptying of the tub.  thus, v = valve1 + valve2+ valve3.\n\nt is unknown, the minutes we are solving for.\n\nx = 1, since we want the time that it takes to fill 1 tub, no more.  putting these all into the equation:\n\n[valve1 + valve2 + valve3]t = 1 tub\n\n[(1/3)tub/min. + (1/5)tub/min. - (2/15)tub/min.]t = 1 tub\n\n[(2/5)tub/min.]t = 1 tub\n\nt = (5/2)min. = 2.5 minutes\n\nchecking the answer, in 2.5 minutes, valve 1 will fill the tub 5/6 of the way, valve 2 will fill it 1/2 of the way, and valve 3 will empty it 1/3 of the way.  (5/6) + (1/2) - (1/3) = 1, therefore our answer works!\n\nt = 2.5min.
451585
water\n\n[edit] 6 seconds too slow!