A conical vessel is 12 feet across the top and 15 feet deep. If it contains a liquid weighing ñ lbs/ft^3 to a depth of 10 feet. Find the work done in pumping the liquid to a height of 3 feet above the vessel.
Is there a number for the desnity?
All I see is the symbol ñ
To find the work done in pumping the liquid to a height of 3 feet above the vessel, we need to calculate the volume of the liquid and then integrate the weight of the liquid over the height difference.
Step 1: Calculate the volume of the liquid
The shape of the vessel is conical. The volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height. In this case, the radius is half of the diameter, so r = 12/2 = 6 feet. The height of the liquid is 10 feet. Therefore, the volume of the liquid is V = (1/3)π(6^2)(10) = 120π cubic feet.
Step 2: Calculate the weight of the liquid
The weight of the liquid is given as ñ lbs/ft^3. Therefore, the weight of the liquid is W = (ñ)(V) = (ñ)(120π) lbs.
Step 3: Calculate the work done in pumping the liquid to a height of 3 feet above the vessel
The work done in lifting an object is given by the formula W = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height lifted. In this case, the mass of the liquid is equal to its weight divided by the acceleration due to gravity, which is approximately 32.2 ft/s^2. So the mass of the liquid is m = W/32.2 lbs. The height lifted is 3 feet. Therefore, the work done is W = (W/32.2)(32.2)(3) = 3(W/32.2) lbs⋅ft.
So, to find the work done in pumping the liquid to a height of 3 feet above the vessel, you need to calculate the weight of the liquid using the given density, and then multiply it by 3/32.2.