Which velocity-time graph matches the position-time graph?
2 answers:
Explanation:
In this question, a position time graph is shown. We need to find the corresponding velocity time graph.
In position time graph, till 1 second the object is in uniform motion and reaches a distance of 30 meters. Then the object is at till 3 seconds. After 3 seconds, the object again reaches a distance of 40 meters uniformly.
Corresponding velocity-time graph is shown in figure (3). The object is moving initially with a speed of 30 m/s. After 1 second, it comes to rest till 3 seconds then it again moves. So, the correct option is (c). Hence, this is the required solution.
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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
