By reading the fine details of the question, carefully and analytically, I have determined that there's no list of modifications to choose from.
The strength of the magnetic field of a solenoid depends on the electric current in its coil windings, the number of wire turns in its coil windings, and the material in its core.
In order to <em>DE</em>crease the strength of its magnetic field, any one or more of these steps could do the job:
-- DEcrease the electric current in its coil windings. This can be accomplished by decreasing the voltage of the power source that energizes the coil, and/or increasing the resistance of the wire in the coil.
-- DEcrease the number of wire turns in the coil.
-- If the solenoid has anything in its core, change the core to something with a lower magnetic 'permeability'. An Iron core will produce the greatest magnetic field strength. Air, vacuum, or NO core will produce the lowest magnetic field strength.
Explanation:
a. KE at bottom = PE at top
½ mv² = mgh
v = √(2gh)
v = √(2 × 9.8 m/s² × 20.0 m)
v = 19.8 m/s
b. Work by friction = PE at top
mgμ d = mgh
d = h / μ
d = 20.0 m / 0.210
d = 95.2 m
Different:
1. number of moons
2. size of rings
3. mass
4. temperature
5. type of rings
i might be wrong though.
This question can be solved from the Kepler's law of planetary motion.
As per this law the square of time period of a planet is proportional to the cube of semi major axis.
Mathematically it can be written as 
⇒
Here K is the proportionality constant.
If
and
are the orbital periods of the planets and
and
are the distance of the planets from the sun, then Kepler's law can be written as-

⇒ 
Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.
Let the distance from sun and orbital period of Saturn is denoted as
and
respectively.
Let the distance from sun and orbital period of earth is denoted as
and
respectively.
we are given that
we know that
1 AU and
1 year.
1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.
Hence distance of Saturn from sun is calculated as -
From Kepler's law as mentioned above-

=![[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU](https://tex.z-dn.net/?f=%5B1%20%5D%5E%7B3%7D%20%2A%5Cfrac%7B%5B29.46%5D%5E%7B2%7D%20%7D%7B%5B1%5D%5E%7B2%7D%20%7D%20AU)

⇒![R_{1} =\sqrt[3]{867.8916}](https://tex.z-dn.net/?f=R_%7B1%7D%20%3D%5Csqrt%5B3%5D%7B867.8916%7D)
=9.5386 AU [ans]
Answer:
Lifting your binder out of your backpack into your top row locker.
Pushing a shopping cart down the aisle at Publix
Explanation:
These are the only two in which an object is moving because of an applied force