A sealed tank 4 feet high with width 3 feet and length 5 feet is filled one foot deep with water. If the tank is turned on one of its smallest faces, how deep (in feet) would the water be?
Please explain how you reached the conclusion?
Thanks
2007-03-26 00:37:48 · 4 answers · asked by thatfella 1 in Science & Mathematics ➔ Mathematics
First, let's figure out the volume of water in the tank.
That's 3 feet * 5 feet * 1 foot, or 15 cubic feet.
Next, let's look at the surface area of the smallest face. That's 3 feet * 4 feet, or 12 square feet.
To find the desired depth, divide 15 cubic feet by 12 square feet. You end up with a depth of 1.25 feet.
2007-03-26 00:42:39 · answer #1 · answered by Bramblyspam 7 · 0⤊ 0⤋
Amount of water in tank
1 x 3 x 5 = 15 cubic feet
The smallest face 3 x 4 is turned down
? x 4 x 3 =15 cubic feet
? x 4 =15/3 sq ft
?= 5/4 ft high 1.25 ft
2007-03-26 07:49:23 · answer #2 · answered by Anonymous · 0⤊ 0⤋
LxWxH=V
3x5x1=15^3ft
3x4xH=15^3
12^2xH=15^3
15^3/12^2=1.25ft
2007-03-26 08:01:17 · answer #3 · answered by Peter H 2 · 0⤊ 0⤋
the height is 1/4 of total. so the volume is also 1/4 of total.
If it sits on any face, just devide the height by 4 to get the answer immediately.
In your case the height is 5 so the height of water will be 1/4 x 5 = 1.25
This way you need not calculate all the time.
2007-03-26 07:50:58 · answer #4 · answered by dipakrashmi 4 · 0⤊ 0⤋