Ahemispherical tank of radius 2 feet is positioned so that its base is circular. how much work is required to fill the tank with water through a hole in the base when the water source is at the base? (the weight-density of water is 62.4 pounds per cubic foot.)

The work required to fill the tank with water through a hole in the base when the water source is at the base is 783.7425 Joules

The formula for calculating the volume of the hemispherical tank is expressed as

The formula for calculating the density is expressed as:

Given the following

Weight-density = 62.4 pounds per cubic foot

From the formula;

The centroid of a hemisphere will be 3/8r above the circular base, so we will raising this weight of water by (3/8)(2 ft) is 3/4 ft

Work done = 1044.99 × 3/4

Workdone = 783.7425 Joules

Hence the work required to fill the tank with water through a hole in the base when the water source is at the base is 783.7425 Joules

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  784 ft·lb

Step-by-step explanation:

The amount of work required is equivalent to the work required to raise the total weight of water to the height of the centroid of the hemisphere.

The weight of the water is ...

  Volume × density = (2/3)πr³ × 62.4 lb/ft³

  = (2/3)π(2 ft)³(62.4 lb/ft³) ≈ 1045.5 lb

A reference page tells us that the centroid of a hemisphere is 3/8r above the circular base, so we're raising this weight of water by (3/8)(2 ft) = 3/4 ft.

The work done is ...

   (3/4 ft)(1045.5 lb) ≈ 784 ft·lb

54 kilometers per hour

step-by-step explanation:

according to rules:

1 km = 1000 meter

which means that 1 meter = 1/1000 km

1 hour = 3600 second [ as 1 hour = 60 min = 60 *60 seconds ]

which means that 1 second = 1/3600 hour

therefore, 1 kilometer/hour =

thus to convert 15 meter per second to kilometer per hour, we multiply 15 to 3.6 kilometer/hour.

15 meter/second=   which is the required answer.