By communication, competition and symbiosis. These are the three main ways organisms interact.
Locations X and Y are at the poles. Hope this helps!
The key to solve this problem is the conservation of momentum. The momentum of an object is defined as the product between the mass and the velocity, and it's usually labelled with the letter
:

The total momentum is the sum of the momentums. The initial situation is the following:

(it's not written explicitly, but I assume that the 5-kg object is still at the beginning).
So, at the beginning, the total momentum is

At the end, we have

(the mass obviously don't change, the new velocity of the 15-kg object is 1, and the velocity of the 5-kg object is unkown)
After the impact, the total momentum is

Since the momentum is preserved, the initial and final momentum must be the same. Set an equation between the initial and final momentum and solve it for
, and you'll have the final velocity of the 5-kg object.
Answer:
16 m/s^2
Explanation:
acceleration tangential = (v^2)/r
a=400/25
a=16 m/s^2
Side note: next time, be more specific when asking about acceleration in circular motion. There's more than one type! Example:
angular acceleration=acceleration tangential/r
angular acc.=16/25
angular acc.=0.64 rad/s^2
Coulomb's Law
Given:
F = 3.0 x 10^-3 Newton
d = 6.0 x 10^2 meters
Q1 = 3.3x 10^-8 Coulombs
k = 9.0 x 10^9 Newton*m^2/Coulombs^2
Required:
Q2 =?
Formula:
F = k • Q1 • Q2 / d²
Solution:
So, to solve for Q2
Q2 = F • d²/ k • Q1
Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9
Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)
Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)
Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)
Then, take the reciprocal of the denominator and start
multiplying
Q2 = 1080 • 1 Coulombs/297
Q2 = 1080 Coulombs / 297
Q2 = 3.63636363636 Coulombs
Q2 = 3.64 Coulumbs