Force (f) = ?
Acceleration (a) = 196 m/s^2
Mass (m) = 0.25 kg
F = (m) • (a)
F = (0.25) • (196)
F = 49 N
Answer : 49 N
I hope that helps you!! Any more questions??
C. 4 N is the answer if not then i be dumb!!!
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)
![\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bq_2%7D%7Bx%5E2%7D%3D%5Cfrac%7Bq_1%7D%7B%282-x%29%5E2%7D)
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.