The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
<h3>Electric field on the master charge</h3>
E = kq/r²
where;
- q is magnitude of master charge
- r is distance of separation
- k is Coulomb's constant
E = (9 x 10⁹ x 0.63)/(0.75²)
E = 1.008 x 10¹⁰ N/C
<h3>Force on the test charge</h3>
F = Eq
where;
- E is electric field
- q is the test charge
F = (1.008 x 10¹⁰) x (0.5)
F = 5.04 x 10⁹ N
Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
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Answer: option D) 42.4 N
The weight of the frame is balanced by the vertical component of tension.
W = T sin θ + T sin θ = 2 T sin θ
The tension in each cable is T = 30 N
Angle made by the cables with the horizontal, θ = 45°
⇒ W = 2×30 N × sin 45° = 2 × 30 N × 0.707 = 42.4 N
Hence, the weight of the frame is 42.4 N. Correct option is D.
Answer:
It says energy can't be created or destroyed
Explanation:
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