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1 year ago

Why is an object’s mass, rather than its weight, used to indicate the amount of matter it contains?.

1 answer:

5 0

An object's mass, rather than its weight is used to indicate the amount of matter it contains because weight is defined as the amount of force due to an existing field (In most cases, gravitational field) that is experienced by the body. The weight of the same body can be different in different environments it's observed.

However, the mass of the body is the exact measure of the amount of matter contained in a body. Which is constant, regardless of the environment, or conditions the body is observed.

Hence, for real-world calculations and experimentation, the mass of the body is considered to represent the amount of matter it contains rather than its weight.

Learn more about mass here

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The situation is represented in the picture attached.

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3 years ago

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A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodie

vivado [14]

Answer:

The length of the rod for the condition on the question to be met is $L = 1.5077 \ m$

Explanation:

The Diagram for this question is gotten from the first uploaded image

From the question we are told that

The mass of the rod is $M = 3.41 \ kg$

The mass of each small bodies is $m = 0.249 \ kg$

The moment of inertia of the three-body system with respect to the described axis is $I = 0.929 \ kg \cdot m^2$

The length of the rod is L

Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

$I = I_r + 2 I_m$

Where $I_r$ is the moment of inertia of the rod about the describe axis which is mathematically represented as

$I_r = \frac{ML^2 }{12}$

And $I_m$ the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as

$I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}$

Thus $2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}$

Hence

$I = M * \frac{L^2}{12} + m * \frac{L^2}{2}$

=> $I = [\frac{M}{12} + \frac{m}{2}] L^2$

substituting vales we have

$0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2$

$L = \sqrt{\frac{0.929}{0.40867} }$

$L = 1.5077 \ m$

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