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2 years ago

7

How can knowing a chemical property such as flammability help you to use a product safely?

1 answer:

aleksandrvk [35]2 years ago

5 0

Answer:

When a substance burns, it produces new substances during a chemical change. Therefore, whether or not a substance is flammable is a chemical property. Knowing which substances are flammable helps you to use them safely. Another chemical property is how compounds react to light

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Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

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