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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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Answer:

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Explanation:

For this problem we will look for the expression for the electric field of a ring of charge, at a point on its central axis.

From the symmetry of the ring, the resulting field is in the direction of the axis, we suppose that it corresponds to the x and y axis and the perpendicular direction but since the two sides of the annulus are offset.

       dE = k dq / r²

The component of this field in the direction to x is

       d Eₓ = dE cos θ

the distance of ring to the point

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where a is the radius of each ring

the cosine is the adjacent leg between the hypotenuse

       Cos θ = x /r

we substitute

   d Ex = k / (x² + a²)  dq      x /√ (x² + a²)

we integrate

      Ex = ∫  k / (x² + a²)^{3/2}  x dq

      Ex = k x/ (x² + a²)^{1.5}   Q

the total field is the sum of the fields created by each ring, since they are on the same line (x-axis), you can perform an algebraic sum

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a) at the origin x = 0

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b) for point x = 40.0cm = 0.400 m

we substitute the values ​​in our equation

       Ex_total = k x [Q (x² + 0.1²)^{1.5} + 2Q / (x² + 0.2²2) ^3/2]

        Ex_total = k Q x [1 / (x² + 0.01)^1.5 + 2 / (x² + 0.04)^1.52]

         

let's calculate

       Ex_total = 9 10⁹ 30 10⁻⁹ 0.40 [1 / (0.4 2 + 0.01) 3/2 + 2 / (0.4 2 + 0.04) 3/2] 10⁻⁹

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3 0
3 years ago
A balloon has a volume of 18.0-L at a pressure of 87.6 kPa. What will be the new volume when the pressure is 48.2 kPa?
Kaylis [27]

Answer:

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Explanation:

We can solve this question using Boyle's law of ideal gas. The law states that the volume and pressure are inversely proportional when the temperature is constant. At first, the volume is 18L (V1) and the pressure is 87.6 kPa(P1). Then the pressure changed into 48.2 kPa(P2). We are asked to find the volume(V2).

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I think

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3 years ago
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as we can write it as

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<u>b. is parallel to the displacement of the object</u>

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garri49 [273]

Answer:

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