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.
Answer:
λ= 5.24 × 10 ⁻² nC/cm
Explanation:
Given:
distance r = 4.10 cm = 0.041 m
Electric field intensity E = 2300 N/C
K = 9 x 10 ⁹ Nm²/C
To find λ = linear charge density = ?
Sol:
we know that E= 2Kλ / r
⇒ λ = -E r/2K (-ve sign show the direction toward the wire)
λ = (- 2300 N/C × 0.041 m) / 2 × 9 x 10 ⁹ Nm²/C
λ = 5.24 × 10 ⁻⁹ C/m
λ = 5.24 nC/m = 5.24 nC/100 cm
λ= 5.24 × 10 ⁻² nC/cm
V = 3.0 x 10⁸ m/s (This is the same for all types of electromagnetic waves)
f = 88.6 MHz = 8.86 x 10⁷ Hz
λ = ?
V = fλ
λ = V/f = (3 x 10⁸)/(8.86 x 10⁷)
= 3.4 m [Ans]
Hope this helps!
The shaft is 78m approximately deep. The correct option is D which is 78 meters
<h3>
What are Sound Waves ?</h3>
Sound waves are longitudinal. That is, the direction of the waves is parallel to the direction of its propagation of particles.
Given that a boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seconds before he hears the "plunk" of the stone hitting the bottom of the shaft.
- The speed of the sound V = 330m/s
- The time it takes the sound to reach the top = t
Speed V = distance / time
330 = h/t
Make t the subject of formula
t = h/330
As the stone is dropped, initial velocity = 0, Using the formula
h = 1/2gT²
But T = 4 - t
T = 4 approximately
Substitute all the parameters
h = 1/2 × 9.8 × (4)²
h = 4.9 × (16)
h = 78.4m
Therefore, the shaft is 78m approximately deep. The correct option is D which is 78 meters
Learn more about sound wave here: brainly.com/question/16093793
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