A tank in the shape of an inverted right circular cone has height 4 meters and radius 2 meters. It is filled with 3 meters of hot chocolate.

Question

A tank in the shape of an inverted right circular cone has height 4 meters and radius 2 meters. It is filled with 3 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is δ=1020 kg/m3.δ=1020 kg/m3. Your answer must include the correct units.

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Linh Đan 2 months 2021-08-02T07:55:59+00:00 1 Answers 3 views 0

Answers ( )

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    2021-08-02T07:57:46+00:00

    Answer:

    The required work to empty the the tank is 247401 J.

    Explanation:

    Given that,

    A tank is in the shape of right circular cone.

    The radius of the tank= 2 m

    The height of the tank = 4 m

    The relation between the radius and the height is

    \frac{r}{h}=\frac{2}{4}

    \Rightarrow r= \frac12 h

    Let h be the height of hot of chocolate any time t.

    The volume of a cone is = \frac 13 \pi r^2 h

                                            =\frac13 \pi (\frac h2)^2h

                                             =\frac16 \pi h^3

    The volume of the chocolate is =\frac16 \pi h^3

    The mass of the  chocolate is(M)= Density × Volume

                                                         =1020(\frac16 \pi h^3) Kg

                                                         =170\pi h^3 kg

    \frac{dM}{dh}= 170\pi (3h^2)

    \Rightarrow dM=510\pi h^2\ dh

    Work done = Force × displacement

                      = Mass × acceleration×displacement

    Here acceleration= acceleration due to gravity = 9.8 m/s²

    The displacement when the hot chocolate level is h is = (4-h)                  

    Work done (dw) =(510\pi h^2 )(9.8)(4-h)dh =4998\pi h^2 (4-h)dh

    Work done = W =\int_0^34998\pi h^2 (4-h)dh

                             =\int_0^34998\pi [4h^2-h^3]dh

                            =4998\pi [4\frac{h^3}{3}-\frac{h^4}{4}]_0^3

                            =4998\pi [(4\frac{3^3}{3}-\frac{3^4}{4})-(4\frac{0^3}{3}-\frac{0^4}{4})]

                            =247401 J

    The required work to empty the the tank is 247401 J.

                           

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