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Physics, 28.06.2019 03:30 Jana1517

An aquarium 4 m long, 1 m wide, and 1 m deep is full of water. find the work needed to pump half of the water out of the aquarium. (use 9.8 m/s2 for g and the fact that the density of water is 1000 kg/m3.)

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Answer from: kayal1076
4900 J

Explanation:

Let's begin by explaining that the Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a distance y.  When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=F.y (1)

In the case of this rectangular aquarium (see figure attached), we know its initial volume V_{i} (which is the same volume occupied by water, because the tank is full) is:

V_{i}=4m.1m.1m=4m^{3} (2)

In addition, we know the force needed to pump a small amount of water is:

F=m_{water}.g (3)

On the other hand, we know the density of the water \rho_{water} is given by the following equation:

\rho_{water}=\frac{m_{water}}{V} (4)

Where m_{water} is the mass of water and V=4m.1m.dy is the volume of a thin "sheet" of water.

Finding m_{water}:

m_{water}=\rho_{water}.V  (5)

Substituting (5) in (3):

F=\rho_{water}.V.g   (6)

And substituting (6) in (1):

W=\rho_{water}.V.g.y (7)

Now, we are asked to find the work needed to pump half of the water out of the aquarium. So, if the aquarium is 1m deep, the half is 0.5m:

W=(1000\frac{kg}{m^{3}})(4m.1m.dy)(9.8\frac{m}{s^{2}})y (8)

W=39200\frac{kg.m^{6}}{s^{2}}ydy (9)

Well, in order to solve this, we have to write the definite integral from y=0 mto y=0.5m:

W=39200\frac{kg.m^{6}}{s^{2}}\int\limits^{0.5}_0 {y} \, dy (10)

Knowing \int\limits^b_a {f(y)} \, dy= F(b)-F(a):

W=39200\frac{kg.m^{6}}{s^{2}}(\frac{(0.5m)^{2}}{2}-\frac{(0m)^{2}}{2}) (11)

W=4900kg\frac{m^{2}}{s^{2}} (12)

W=4900J >>>>This is the work needed to pump half of the water out of the aquarium


An aquarium 4 m long, 1 m wide, and 1 m deep is full of water. find the work needed to pump half of
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Answer from: Quest

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!

ansver
Answer from: Quest

the answer should be in the picture.

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