(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.
Recall that

We have
, so that

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.
(c) The ball's average velocity is 0. Average velocity is given by
, and we know that
.
(d) The position of the ball
at time
is given by

Take the starting position to be the origin,
. Then after 6 seconds,

so the ball is 42 m away from where it started.
We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

Since the velocity is positive, the ball is still moving up the incline.
Answer: b) they are the areas where Earth's magnetic field is weakest
Explanation:
According to classical physics, a magnetic field always has two associated magnetic poles (north and south), the same happens with magnets. This is because for <em>classical physics</em>, naturally, magnetic monopoles can not exist.
In this context, Earth is similar to a magnetic bar with a north pole and a south pole. This means, the axis that crosses the Earth from pole to pole is like a big magnet.
Now, by convention, on all magnets the north pole is where the magnetic lines of force leave the magnet and the south pole is where the magnetic lines of force enter the magnet. Then, for the case of the Earth, the north pole of the magnet is located towards the geographic south pole and the south pole of the magnet is near the geographic north pole.
Being the magnetic poles the places where the Earth's magnetic field is weakest. And it is for this reason, moreover, that the magnetic field lines enter the Earth through its magnetic south pole (which is the geographic north pole).
"did you press the switch?"
Answer:
The equivalent resistance of the parallel circuit would be 20 Ω
Explanation:
To calculate the resistance of resistors connected in parallel, the formula to be used is
1/R = 1/R₁ + R₂ + R₃ + R₄...
1/R = 1/120 + 1/60 + 1/40
1/R = (1 + 2 + 3)/120
1/R = 6/120
1/R = 1/20 Ω
This can be rewritten or cross-multiplied to be
R × 1 = 20 × 1
R = 20 Ω
The equivalent resistance (R) would then be 20 Ω
I think it’s the first one