The current passing through a circuit consisting of a battery of 12 V and resistor of 2 ohms is 6 Ampere
.
Explanation:
- Assume the wires are ideal with zero resistance.
- The current passing through the circuit will be
I = V/R = 12/2 = 6.000 A.
Answer:
the energy of the spring at the start is 400 J.
Explanation:
Given;
mass of the box, m = 8.0 kg
final speed of the box, v = 10 m/s
Apply the principle of conservation of energy to determine the energy of the spring at the start;
Final Kinetic energy of the box = initial elastic potential energy of the spring
K.E = Ux
¹/₂mv² = Ux
¹/₂ x 8 x 10² = Ux
400 J = Ux
Therefore, the energy of the spring at the start is 400 J.
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