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A uniform solid consists of a hemisphere with centre $O$ and radius $0.6{\text{ }}m$ joined to a cylinder of radius $0.6{\text{ }}m$ and height $0.6{\text{ }}m$. The plane face of the hemisphere coincides with one of the plane faces of the cylinder.

a) Calculate the distance of the centre of mass of the solid from $O$.

[The volume of a hemisphere of radius $r$ is $\frac{2}{3}\pi {r^3}$]

b) A cylindrical hole, of length $0.48{\text{ }}m$, starting at the plane face of the solid, is made along the axis of symmetry (see diagram). The resulting solid has its centre of mass at $O$. Show that the area of the cross-section of the hole is $\frac{3}{{16}}\pi {m^2}$.

c) It is possible to increase the length of the cylindrical hole so that the solid still has its centre of mass at $O$. State the increase in the length of the hole.

پاسخ تشریحی :
نمایش پاسخ

a) $\pi {0.6^2} \times 0.6 \times 0.3 - 2\pi {\text{ }}{0.6^3}/3 \times 3 \times 0.6/8$

$ = \left( {\pi {{0.6}^3} + 2\pi {{0.6}^3}/3} \right)d$

$d = 0.09{\text{ }}m$

b) $\frac{2}{3}\pi {0.6^3} \times \frac{3}{8} \times 0.6 - \pi  \times {0.6^3} \times 0.3$

$ + 0.48A \times 0.36 = 0$

$A = 3\pi /16{\text{ }}{m^2}$

OR

$[\frac{2}{3}\pi  \times {0.6^3} + \pi  \times {0.6^3}] \times 0.09 = 0.48A \times 0.36$

$A = 3\pi /16$

c) Increase in length $\left[ { = 2 \times \left( {0.6 - 0.48} \right)} \right] = 0.24m$

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