What would be new resistance if length of conductor is doubled and thickness is halved
1 answer:
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Answer:
a) k = 200 N/m
b) E = 4 J
c) Δx = 6.3 cm
Explanation:
a)
- In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:
- where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
- Solving for k:
b)
- Assuming no friction present, total mechanical energy mus keep constant.
- When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:
- Replacing k and Δx by their values, we get:
c)
- When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
- We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
- So, we can write the following expression:
- Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:
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Answer:
(a) Elongation of rod = 0.19732 mm
(b) Change in diameter = 0.00651 mm
Explanation:
Circular area at end of steel rod = pi * diameter^2 / 4
Area = 3.801 * 10^(-4) meter squared
Stress = force / area
Stress = 75000 / (3.801 * 10^-4)
Stress = 197316495 Pa OR 0.197 GPa
Modulus of elasticity = stress / strain
200 = 0.19732 / Strain
Strain = 0.0009866 (longitudinal)
(a) Strain = change in length / initial length
0.0009866 = Elongation of rod / 200
Elongation of rod = 0.19732 mm
(b) Poisson ratio = lateral strain / longitudinal strain
0.3 = lateral strain / 0.0009866
Lateral strain = 0.000296
Lateral strain = Change in diameter / original diameter
0.000296 = Change in diameter / 22
Change in diameter = 0.00651 mm