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V125BC [204]
4 years ago
8

Three importance of SI system​

Physics
1 answer:
diamong [38]4 years ago
8 0

Answer:

Firstly they are, by design, easy to use in most scientific and engineering calculations; you only ever have to consider multiples of 10. If I’m given a measurement of 3.4 kilometres, I can instantly see that it’s 3′400 metres, or 0.0034 Megametres, or 3′400′000 millimetres. It’s not even necessary to use arithmetic, I just have to remember the definitions of the prefixes (“kilo” is a thousand, “megametre” is a million, “milli” is a thousandth) and shift the decimal point across to the left or the right. This is especially useful when we’re considering areas, speeds, energies, or other things that have multiple units; for instance,

1 metre^2 = (1000millimetre)^2 = 1000000 mm^2.

If we were to do an equivalent conversion in Imperial, we would have

1 mile^2 = (1760 yards)^2

and we immediately have to figure out what the square of 1760 is! However, the fact that SI is based on multiples of 10 has the downside that we can’t consider division by 3, 4, 8, or 12 very easily.

Secondly they are (mostly) defined in terms of things that are (or, that we believe to be) fundamental constants. The second is defined by a certain kind of radiation that comes from a caesium atom. The metre is defined in terms of the second and the speed of light. The kelvin is defined in terms of the triple point of water. The mole is the number of atoms in 12 grams of carbon-12. The candela is defined in terms of the light intensity you get from a very specific light source. The ampere is defined using the Lorentz force between two wires. The only exception is the kilogram, which is still defined by the mass of a very specific lump of metal in a vault in France (we’re still working on a good definition for that one).

Thirdly, most of the Imperial and US customary units are defined in terms of SI. Even if you’re not personally using SI, you are probably using equipment that was designed using SI.

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Which of the following changes would double the force between two charged particles? O A. Decreasing the distance between the pa
jarptica [38.1K]

Let's check

\\ \rm\dashrightarrow F=\dfrac{k}{q_1q_2}{r^2}

\\ \rm\dashrightarrow F\propto Q

\\ \rm\dashrightarrow F\propto \dfrac{1}{r^2}

So

Option A and C can be used

6 0
2 years ago
Read 2 more answers
For each combination, record the total force, total mass, and acceleration. On the last column, multiply the mass (m) and accele
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Hello. You did not present the combinations the question refers to, which makes it impossible for this question to be answered accurately. However, I will try to help you in the best possible way.

To present the total force you must use the following formula: Mass x Acceleration.

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8 0
3 years ago
A 10 kg box accelerates for at a rate of 12 miles per second what is the force acting on the box
lions [1.4K]

193080N

Explanation:

Given parameters:

Mass of the box = 10kg

Acceleration = 12miles per second

Unknown:

Force on the box = ?

Solution:

The force acting on the box is function of the mass and acceleration of the box.

  Force = mass x acceleration

   Now we need to convert the given acceleration to m/s

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    \frac{12 miles }{seconds } x \frac{1.609km}{1miles} x \frac{1000}{1km}

      = 19308m/s

 Force on box = 10 x 19308 = 193080N

Learn more:

Force problems brainly.com/question/4033012

#learnwithBrainly

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