r₁ = distance of point A from charge q₁ = 0.13 m
r₂ = distance of point A from charge q₂ = 0.24 m
r₃ = distance of point A from charge q₃ = 0.13 m
Electric field by charge q₁ at A is given as
E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C towards right
Electric field by charge q₂ at A is given as
E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C towards left
Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction
Electric field at A will be zero when
E₁ = E₂ + E₃
1.225 = 0.703 + E₃
E₃ = 0.522 N/C
Electric field by charge "q₃" is given as
E₃ = k q₃ /r₃²
0.522 = (9 x 10⁹) q₃/(0.13)²
q₃ = 0.980 x 10⁻¹² C = 0.980 pC
Answer:
The voltage induced in the coil is 1.25 V.
Explanation:
Given;
number of turns, N = 100 turns
cross sectional area of the copper coil, A = 0.1 m²
initial magnetic field, B₁ = 0.5 T
final magnetic field, B₂ = 1.00 T
duration of change in magnetic field, dt = 4 s
The induced emf in the coil is calculated as;
Therefore, the voltage induced in the coil is 1.25 V.
Answer:a)45cos40 b)45sin40 c) 2.95 d)102
Explanation:
a)
b)
c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -<em>g</em>. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: .
d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.