Answer:
It is a force in the direction of the motion that allows the box to move
Explanation:
<span>Social
i think so ,but i am not sure</span>
The correct answer is C. Final Velocity
Hope this helped!
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.
Answer:
The hollow cylinder rolled up the inclined plane by 1.91 m
Explanation:
From the principle of conservation of mechanical energy, total kinetic energy = total potential energy
The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.
moment of inertia, I, of a hollow cylinder = ¹/₂mr²
substitute for I in the equation above;
given;
v₁ = 5.0 m/s
vf = 0
g = 9.8 m/s²
Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m