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Ksivusya [100]
3 years ago
6

Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.

Engineering
1 answer:
Sladkaya [172]3 years ago
5 0

Answer:

Minimum area of rectangle = 24 cm²

Explanation:

Given:

Length of rectangle = 6 cm

Width of rectangle = 4 cm

Find:

Minimum area of rectangle

Computation:

Area of rectangle = Length of rectangle x Width of rectangle

Minimum area of rectangle = Length of rectangle x Width of rectangle

Minimum area of rectangle = 6 x 4

Minimum area of rectangle = 24 cm²

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Solution:

Given:

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Coulombian force for two point charges seperated by a distance 'r' is given by:

F = \frac{1}{4\pi\epsilon _{_{o}} }\times \frac{q_{1}q_{2}}{r^{2}}      (1)

Now, we know that

\frac{1}{4\pi \epsilon _{o}} = 9\times 10^{^{9}}

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where, n = no. of electrons

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q_{1} = n_{1}e = 2e

q_{2} = n_{2}e = 2e

Using Eqn (1)

F = (9\times 10^{^{9}})\times \frac{4\times (1.6\times 10^{-19})^{^{2}}}{(1.25\times 10^{-9})^{2}}

F = 5.89\times10^{-10} N

Since, the force is between two opposite charged ions, it is attractive in nature.

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