1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.
2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.
3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.
Speed v = initial speed u + acceleration a x time t
v=u+at = 2 + 4*3 = 14 m/s
When you breathe on a cold window and it fogs up, that change of state is called: Condensation.
Condensation is: The process in which molecules of a gas slow down, come together, and form a liquid water that collects as droplets on a cold surface when humid air is in contact with it.
I know this because I learned it in science class in school.
Hope I helped!
- Debbie
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)

q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.