We apply the formula S=½gt², where S represents the vertical distance of 74.3m, and g is the acceleration due to gravity valued at 9.8 m/s². Therefore, t=√(2S/g)=3.884 seconds. Consequently, the car remains airborne for around 3.894 seconds. Its horizontal journey is vt=20×3.894=77.88m.
Response:
Clarification:
The force between two charges, q₁ and q₂ at a distance d is represented by the formula
F = k q₁ q₂ / d²
Here, the force between charge q₁ = -15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C with distance d = (1.66 - 1.24) = 0.42 mm
k = 1 / (4π x 8.85 x 10⁻¹²)
Substituting the values into the equation
F = 1 / (4π x 8.85 x 10⁻¹²) x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²
= 9 x 10⁹ x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²
= 35969.4 x 10⁻³ N.
For the force between charge q₂ = 34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C at a distance d = (1.24 - 0) = 1.24 mm.
Substituting the values into the expression
F = 1 / (4π x 8.85 x 10⁻¹²) x 34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²
= 9 x 10⁹ x -34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²
= 82729.6 x 10⁻³ N
Both forces direct towards the left (away from the origin, towards the negative x-axis)
Total force = 118699 x 10⁻³
= 118.7 N.
