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irina1246 [14]
2 years ago
10

A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?

Physics
2 answers:
jekas [21]2 years ago
7 0

A reference point would be something not on the ship which could be used to calculate distance traveled.

Answer: C.) A lighthouse on a nearby Island

Alexxx [7]2 years ago
5 0

Answer:

C.) A lighthouse on a nearby Island

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Using the formula

a\: = \frac{v - u}{t}

a = \frac{35 m/s \: - \: 22 m/s}{5s}

a = 2.6 {m/s^2}
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is dimensionally correct relation necessarily to be a correct physical relation? explain with example.​
Andreas93 [3]

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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