Force applied on the car due to engine is given as
towards right
Also there is a force on the car towards left due to air drag
towards left
now the net force on the car will be given as
now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.
So we can say
So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.
Answer:
The third charged particle must be placed at x = 0.458 m = 45.8 cm
Explanation:
To solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁, q₂: Charges in Coulombs (C)
d: distance between the charges in meters (m)
Equivalence
1μC= 10⁻⁶C
1m = 100 cm
Data
K = 8.99 * 10⁹ N*m²/C²
q₁ = +5.00 μC = +5.00 * 10⁻⁶ C
q₂= +7.00 μC = +7.00 * 10⁻⁶ C
d₁ = x (m)
d₂ = 1-x (m)
Problem development
Look at the attached graphic.
We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:
We use formula (1) to calculate the forces F₁₃ and F₂₃
F₁₃ = F₂₃
We eliminate k and q₃ on both sides
We eliminate 10⁻⁶ on both sides
We solve the quadratic equation:
In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .
x = 0.458m = 45.8cm
