"Fluid intelligence involves being able to think and reason abstractly and solve problems. This ability is considered independent of learning, experience, and education. Examples of the use of fluid intelligence include solving puzzles and coming up with problem-solving strategies."
- Verywell Mind
Answer:
The speed it reaches the bottom is

Explanation:
Given:
, 
Using the conservation of energy theorem


, 
![m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2](https://tex.z-dn.net/?f=m%2Ag%2Ah%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2A%28r%2Aw%29%5E2%20%2B%5Cfrac%7B1%7D%7B2%7D%2A%5B%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Ar%5E2%5D%2Aw%5E2)


Solve to w'





This question can be solved by using the equations of motion.
a) The initial speed of the arrow is was "9.81 m/s".
b) It took the arrow "1.13 s" to reach a height of 17.5 m.
a)
We will use the second equation of motion to find out the initial speed of the arrow.

where,
vi = initial speed = ?
h = height = 35 m
t = time interval = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,

<u>vi = 9.81 m/s</u>
b)
To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 17.5 m
vi = initial speed = 9.81 m/s
t = time = ?
Therefore,

solving this quadratic equation using the quadratic formula, we get:
t = -3.13 s (OR) t = 1.13 s
Since time can not have a negative value.
Therefore,
<u>t = 1.13 s</u>
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse
To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.
The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,


M,m = Counts per second
Our radios are given by



Therefore replacing we have that,






Therefore the number of counts expect at a distance of 20 cm is 19.66cps