The correct answer is the Black Market.
The Black Market refers to the informal (and sometimes formal) system of trading of illegal goods by merchants and dealers of products such a weapons, drugs, and good that have been banned by countries. This can also apply to informal marketplaces in countries with sanctions placed upon them get goods that formal treaties are blocking the people from accessing.
Answer:
4 kg → +4 m/s
5 kg → -5 m/s
Explanation:
The law of conservation of momentum states that:
- m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
- left side → velocities before collision
- right side → velocities after collision
You'll notice that we have two missing variables: v₁' & v₂'. Assuming this is a perfectly elastic collision, we can use the conservation of kinetic energy to set the initial and final velocities of the individual bodies equal to each other.
Let's substitute all known variables into the first equation.
- (4)(-6) + (5)(3) = (4)v₁' + (5)v₂'
- -24 + 15 = 4v₁' + 5v₂'
- -9 = 4v₁' + 5v₂'
Let's substitute the known variables into the second equation.
- (-6) + v₁' = (3) + v₂'
- -9 = -v₁' + v₂'
- 9 = v₁' - v₂'
Now we have a system of equations where we can solve for v₁ and v₂.
- -9 = 4v₁' + 5v₂'
- 9 = v₁' - v₂'
Use the elimination method and multiply the bottom equation by -4.
- -9 = 4v₁' + 5v₂'
- -36 = -4v₁' + 4v₂'
Add the equations together.
<u>The final velocity of the second body (5 kg) is -5 m/s</u>. Substitute this value into one of the equations in the system to find v₁.
- 9 = v₁' - v₂'
- 9 = v₁' - (-5)
- 9 = v₁' + 5
- 4 = v₁'
<u>The final velocity of the first body (4 kg) is 4 m/s.</u>
<u></u>
We can verify our answer by making sure that the law of conservation of momentum is followed.
- m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
- (4)(-6) + (5)(3) = (4)(4) + (5)(-5)
- -24 + 15 = 16 - 25
- -9 = -9
The combined momentum of the bodies before the collision is equal to the combined momentum of the bodies after the collision. [✓]
Answer:
The point at which the electrical potential is zero is x = +0.33 m.
Explanation:
By definition the electrical potential is:
![V_{E} = \frac{K*q}{r}](https://tex.z-dn.net/?f=%20V_%7BE%7D%20%3D%20%5Cfrac%7BK%2Aq%7D%7Br%7D%20)
Where:
K: is Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
r: is the distance
The point at which the electrical potential is zero can be calculated as follows:
![V_{1} + V_{2} = 0](https://tex.z-dn.net/?f=%20V_%7B1%7D%20%2B%20V_%7B2%7D%20%3D%200%20)
(1)
q₁ is the first charge = +3 mC
r₁ is the distance from the point to the first charge
q₂ is the first charge = -6 mC
r₂ is the distance from the point to the second charge
By replacing r₁ = 1 - r₂ into equation (1) we have:
(2)
By solving equation (2) for r₂:
![r_{2} = \frac{q_{1}}{q_{1} - q_{2}} = \frac{3 mC}{3 mC - (-6 mC)} = +0.33 m](https://tex.z-dn.net/?f=r_%7B2%7D%20%3D%20%5Cfrac%7Bq_%7B1%7D%7D%7Bq_%7B1%7D%20-%20q_%7B2%7D%7D%20%3D%20%5Cfrac%7B3%20mC%7D%7B3%20mC%20-%20%28-6%20mC%29%7D%20%3D%20%2B0.33%20m)
Therefore, the point at which the electrical potential is zero is x = +0.33 m.
I hope it helps you!
Unless the second ball's speed is twice the amount of the first ball's no they will not cross paths