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.
1) 
2) 13.9 s
Explanation:
1)
The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.
It can be calculated using the equation:
(1)
where
G is the gravitational constant
is the Earth's mass
r is the distance of the object from the Earth's center
The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is
where
is the Earth's radius
Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):
2)
The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula
where
L is the length of the pendulum
g is the acceleration due to gravity at the location of the pendulum
Note that the period of a pendulum does not depend on its mass.
For the pendulum in this problem, we have:
L = 3 m is its length
is the acceleration due to gravity (calculated in part 1)
Therefore, the period of the pendulum is:
Answer:
F = 294.3 [N]
Explanation:
To solve this problem we must use Newton's second law which tells us that force is equal to the product of mass by acceleration. It is this particular case the acceleration is due to the gravitational acceleration since the body is in free fall.
Therefore we have:
F = m*g
where:
F = force [N]
m = mass = 30 [kg]
g = gravity acceleration = 9.81 [m/s^2]
F = 30*9.81
F = 294.3 [N]
Density=m/v so d=22.7/2 which gives you an answer of 11.35
At the bottom of the first and biggest drop. I hope I helped ^u^
