Physics, 29.01.2021 01:00 24chrim

A point charge, Q1 = 12.0 C, is placed at the origin (0 cm, 0 cm) and a second charge, Q2, is placed at the coordinates (4.00 cm, 0 cm). A third charge, Q3 = 15.00 C, is placed at (5.0 cm, 0 cm). The force on Q3 is F⃑ = −20.0 N î. What is the value and sign of Q2?

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Answer from: mv603177

q₂ = -4.80 10⁻⁴ C = - 0.48 mC, charge is negative

Explanation:

Let's use coulomb's law

F = $k \frac{q_1 q_2}{r^2}$

and the sum of forces, remember that charges of the same sign repel and of different sign attract

∑ F = F₁₃ + F₂₃ (1)

Let's start by fixing a reference system located at charge 1 with the positive direction to the right. In the problem it indicates that the net force on charge 3 is F = - 20.0 N, the negative sign indicates that the force is towards the left

let's look for every force, the charge q₁ = 12 10-⁻³ C and q₃ = 15 10⁻³ C

F₁₃ = $k \frac{q_1 q_3}{x_{13}^2}$

F₁₃ = 9 10⁹ 12.0 15.0 10⁻⁶ / (5-0)²

F₁₃ = 64.8 10 3 N

This force is repulsive, that is, it is directed to the right

F₂₃ = k \frac{q_2 q_3}{x_{23}^2}

F₂₃ = 9 10⁹ q₂ 15.0 10⁻³ / (5-4)²

F₂₃ = 135 10⁶ q₂ N

we substitute in equation 1

-20.0 = 64.8 10³ + 135 10⁶ q₂

q₂ = (-20 - 64.8 10³) / 135 10⁶

q₂ = -4.80 10⁻⁴ C

the sign indicates that the charge is negative

Answer from: Quest

19 hertz bruh u trippin? ?

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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A point charge, Q1 = 12.0 C, is placed at the origin (0 cm, 0 cm) and a second charge, Q2, is place...