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pochemuha
2 years ago
10

A block oscillating on a spring has an amplitude of 20 cmcm. Part A Part complete What will the block's amplitude be if its tota

l energy is doubled
Physics
1 answer:
Gnom [1K]2 years ago
7 0

Hi there!

Recall that the amplitude of an oscillating system occurs at a point of maximum potential energy.

For a spring system, potential energy is given as:
U = \frac{1}{2}kx^2

U = Potential Energy (J)

k = Spring constant (N/m)
x = amplitude (m)

We can rearrange the equation to solve for amplitude more easily.

U = \frac{1}{2}kx^2\\\\2U = kx^2\\\\x^2 = \frac{2U}{k}\\\\x = \sqrt{\frac{2U}{k}}

If we double 'U':
x' = \sqrt{\frac{2(2U)}{k}} = \sqrt{2}*\sqrt{\frac{2U}{k}}\\\\x' = \sqrt{2} * x

Thus, the new amplitude would be √2 times greater, so:
20 cm \cdot \sqrt{2} = \boxed{28.284 cm}

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A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
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Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

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