An oil tank is the shape of an inverted right circular cone with the pointed end down. The tank is 15 feet tall and is 12 feet in diameter at the top. At 1:00pm, the tank has oil 5 feet deep in it. Oil is pouring in at 5 cubic feet per minute. To the nearest minutes, at what time will the tank be full?
v = 1/3 pi r^2 h
= 1/3 pi * 36 * 15 = 565.487 ft^3
The volume already full at 1:00 is 62.832, leaving 502.655 ft^3
502.655/5 = 100.531 min. = 1hr 41min
so, tank is full at 2:41 pm
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To find the time at which the tank will be full, we need to determine the rate at which the oil level in the tank is rising.
First, let's calculate the volume of the oil in the tank when the oil is 5 feet deep. Since the oil tank is in the shape of an inverted right circular cone, we can use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Where V is the volume, π is the mathematical constant pi (approximately 3.14159), r is the radius of the top of the tank (half the diameter), and h is the height of the oil in the tank.
Given:
Top diameter = 12 feet
Radius (r) = 12/2 = 6 feet
Height (h) = 5 feet
V = (1/3) * π * 6^2 * 5
V ≈ 188.5 cubic feet
Now, let's calculate the time it takes to fill the remaining volume, which is the difference between the full volume of the tank and the volume when the oil is 5 feet deep.
Full volume = (1/3) * π * r^2 * h
Full volume ≈ (1/3) * π * 6^2 * 15
Full volume ≈ 565.5 cubic feet
Remaining volume = Full volume - Volume at 5 feet
Remaining volume = 565.5 - 188.5
Remaining volume ≈ 377 cubic feet
Since the oil is pouring into the tank at a rate of 5 cubic feet per minute, we can divide the remaining volume by the pouring rate to find the time it takes to fill the remaining volume.
Time = Remaining volume / Pouring rate
Time ≈ 377 / 5
Time ≈ 75.4 minutes
Therefore, to the nearest minute, the tank will be full at approximately 1:00 pm + 75 minutes, which is 2:15 pm.