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'





The equation
(option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:


Where:
- m₁: is the mass of the lab cart = 15 kg
- m₂: is the <em>mass </em>of the object dropped = 2 kg
: is the initial velocity of the<em> lab cart </em>
: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
: is the final velocity of the<em> lab cart </em>
: is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

Therefore, the equation
represents the horizontal momentum (option 3).
Learn more about linear momentum here:
I hope it helps you!
Answer:
14m/s
Explanation:
Given parameters:
Radius of the curve = 50m
Centripetal acceleration = 3.92m/s²
Unknown:
Speed needed to keep the car on the curve = ?
Solution:
The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.
It is given as;
a =
a is the centripetal acceleration
v is the speed
r is the radius
Now insert the parameters and find v;
v² = ar
v² = 3.92 x 50 = 196
v = √196 = 14m/s