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Answer:
The magnitude of the sum of the two vectors is approximately 169.34 m
Explanation:
The given vectors are;
Vector B;
Magnitude = 101 m
Direction = 60.0° (30.0° west of north)
Resolving the vector into its x and y component vectors gives;
= 101 × cos(60.0°)·i + 101 × sin(60.0°)·j = 55·i + 55·√3·j
∴ = 55·i + 55·√3·j
Vector A;
Magnitude = 85.0 m
Direction = West
Resolving the vector into its x and y component vectors gives;
= 85.0 × cos(0.0°)·i + 85.0 × sin(0.0°)·j = 85.0·i + 0·j
∴ = 85.0·i
The sum of the two vectors is +
= 55·i + 55·√3·j + 85.0·i = 140.0·i + 55·√3·j
The direction of the sum of the two vectors, θ = arctan(y-component of the vector)/(x-component of the vector))
Therefore, θ = arctan((55·√3)/140) ≈ 34.23° north of west or 55.77° west of north
The magnitude of the sum of the two vectors, R is R = √(140² + (55·√3)²) ≈ 169.34 m.
Answer:
Following are the solution to the given question:
Explanation:
For charging plates that are connected in a similar manner:
Calculating the total charge:
Calculating the common potential:
Calculating the charge after redistribution: