Let's break the question into two parts:
1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.
1. Ramp Scenario: In an incline, the only component of cart's weight(
mg) that is in the direction of motion is
. Therefore the effort force in this case must be equal or greater than
.
Now we need to find

.

is the angle between the incline of the ramp and the ground.
Since the height is
5m and the length of the ramp is
8m, 
would be
5/8 or 0.625. Now that you have

, mutiple it with
mg.
=> m*g*

= 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is
125N.
2. Lever Scenario:
Just apply "moment action" in this case, which is:


= ?

= mg = 20 * 10 = 200N

= 10m

= 1m
Plug-in the values in the above equation:

= 200/10=
20NAs 20N << 125N, the best choice is to use lever.
Answer:
The slope is 0.5
Explanation:
It increases so it is positive and it increases by half a block each time so 0.5.
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Answer:
1. Velocity = 75.39 m/s
2. Air resistance
3. Mass of the coin
Explanation:
1. Given that the height of the building is 290 m. From Newton's third law of motion,
=
+ 2gh
where: V is the final velocity of the coin, U is the initial velocity, g is the force of gravity and h is the height.
Since the coin was dropped from a height, its initial velocity is zero. The force of gravity is taken as 9.8 m/
, so that:
= 0 + 2 x 9.8 x 290
= 5684
V = 
= 75.3923
The velocity with which the coin hits the ground is 75.39 m/s.
2. Air resistance: During the free fall of the coin, the direction of wind flow could either cause an increase or decrease the value predicted.
ii. Mass of the coin: This can also affect the predicted value.