is dimensionally correct relation necessarily to be a correct physical relation? explain with example.

Physics

1 answer:

Andreas93 [3]3 years ago

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Answer: hope it helps you...❤❤❤❤

Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.

No, not necessarily.

For instance, Newton’s 2nd law is F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.

But take a look at this (incorrect) equation for the force of gravity:

F=−G(m+M)Mm√|r|3r

It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.

It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.

A simpler counter example is 1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.

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A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,

Marina CMI [18]

Answer:

B. Attract each other with a force of 10 newtons.

Explanation:

Statement is incorrectly written. <em>The correct form is: A </em> $+10\,\mu C$ <em> charge and a </em> $-10\,\mu C$ <em> at a distance of 0.3 meters. </em>

The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$ (1)

Where:

$F$ - Electrostatic force, in newtons.

$\kappa$ - Electrostatic constant, in newton-square meters per square coulomb.

$|q_{A}|,|q_{B}|$ - Magnitudes of electric charges, in coulombs.

$r$ - Distance between charges, in meters.

If we know that $\kappa = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $|q_{A}| = |q_{B}| = 10\times 10^{-6}\,C$ and $r = 0.3\,m$ , then the magnitude of the electrostatic force is: