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nexus9112 [7]
3 years ago
10

a uniform disc and hollow right circular cone have the same formula for their moment of inertia when rotating about the central

axis why is it so?
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
6 0

Answer:

This is as a result that about the central axis a collapsed hollow cone is equivalent to a uniform disc

Explanation:

The integration of the differential mass of the hollow right circular cone yields

I=\int\limits   dmr^2 = \int\limits^a_b {\frac{2Mxr^2}{R^2 +H^2} } \, dx  = \frac{2MR^2dx}{(R^2 +H^2)^2} \frac{(R^2 +H^2)^2}{4} = \frac{1}{2}MR^2

and for a uniform disc

I = 1/2πρtr⁴ = 1/2Mr².

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igomit [66]

Answer:

T=Lnsin\alpha

Please check the attached

Explanation:

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A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is
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(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

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- the buoyant force, equal to B=m_w g (m_w=mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

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m'g = mg-m_w g

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(b) 1.98\cdot 10^{-4} m^3

If the rock is completely submerged, the volume of the rock corresponds to the volume of water  displaced.

The volume of water displaced is given by

V_w = \frac{m_w}{\rho_w}

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m_w = 198 g = 0.198 kg is the mass of the water displaced

\rho_w = 1000 kg/m^3 is the density of the water

Substituting,

V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3

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(c) 2727 kg/m^3

The average density of the rock is given by

\rho = \frac{m}{V}

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m = 540 g = 0.540 kg is the mass of the rock

V=1.98\cdot 10^{-4} m^3 is its volume

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\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3

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