Answer :

mstepffer

Answer:

The work W required to pump the oir out of the spout is [tex]W=62344.9 J[/tex]

Step-by-step explanation:

We are given the density of the oil, and we can calculate the volume of the oil as

[tex]V=\pi r^2\frac{h}{2}[/tex]

because the tank is half full.

The we can calculate the oil mass as

[tex]m=\rho V=\rho \pi r^2\frac{h}{2}[/tex]

Now, to calculate the work W necessary to pump the oil out, we have that

[tex]W=\int\limits^{h_{f}} _{h_{i}} {mg} \, dh[/tex]

where m is already calculated, g is given in the problem, and h goes from 0m to 1m.

Finally we can calculate the work W:

[tex]W=\int\limits^{1m} _{0m} {\rho g \pi r^2 \frac{h}{2}} \, dh=\frac{\rho}{2} g \pi r^2\int\limits^{1m} _{0m} {h} \, dh[/tex]

Therefore

[tex]W=\frac{\rho}{4} g \pi r^2 m^2=\frac{900}{4}\frac{kg}{m^3}*9.8\frac{m}{s^2}*9\pi m^2*m^2=19845\pi Nm=62344.9 J[/tex]

is the work required to pump the oil out of the spout.

raphealnwobi

The work required to pump the oil out of the spout is 4 × 10⁶ Joules.

Using Pythagoras theorem gives:

[tex]r^2+y^2=3^2\\\\r^2=9-y^2\\\\r=\sqrt{9-y^2} \\\\\\The\ volume(V)\ is:\\\\V=\pi r^2\Delta y\\\\V=\pi (\sqrt{9-y^2})^2\Delta y\\\\V=\pi(9-y^2)\Delta y\\\\\\Mass(m)=density*volume\\\\m=\pi(9-y^2)\Delta y*900\\\\m=900\pi(9-y^2)\Delta y\\\\\\Force=mass*acceleration\ due \ to\ gravity\\\\Force=900\pi(9-y^2)\Delta y*9.8\\\\Force=8820\pi(9-y^2)\Delta y\\\\\\A\ distance\ of\ 4-y\ is\ moved,hence:\\\\work=force*distance\\\\work(W)=8820\pi(9-y^2)\Delta y*(4-y)\\\\[/tex]

[tex]work(W)=8820\pi(9-y^2)\Delta y*(4-y)\\\\W=\int\limits^3_{-3} {8820\pi(9-y^2)(4-y)} \, dy\\\\W=4*10^6\ J[/tex]

The work required to pump the oil out of the spout is 4 × 10⁶ Joules.

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