The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316
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Answer:
2.12/R mW
Explanation:
The electrical power, P generated by the rod is
P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?
So, the induced emf on the conductor is
E = BLv
= 0.575 T × 0.2 m × 0.4 m/s
= 0.046 V
= 46 mV
The electrical power, P generated by the rod is
P = B²L²v²/R
= B²L²v²/R
So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²
= 0.002116/R W
= 2.12/R mW
Answer:
Find the set of value of x Find the set of value of x which satisfy the inequality 2r2- 5x 2 18which satisfy the inequality 2r2- Find the set of value of x which satisfy the inequality 2r2- 5x 2 185x 2 18
