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Lina20 [59]
3 years ago
15

LASIK eye surgery uses pulses of laser light to shave off tissue from the cornea, reshaping it. A typical LASIK laser emits a 1.

0-mm-diameter laser beam with a wavelength of 193 nm. Each laser pulse lasts 20 ns and contains 2.0 mJ of light energy.
Part A: What is the power of one laser pulse?
Part B: During the very brief time of the pulse, what is the intensity of the light wave?
Physics
1 answer:
pickupchik [31]3 years ago
6 0

Answer:

a)     P = 1.0 10⁵ W, b)   I = 1.27 10¹¹ W / m²

Explanation:

a) the power is defined by the unit of time work

       P = W / t

Work is equal to the variation of energy

       P = E / t

Let's calculate

      P = 2.0 10⁻³ / 20 10⁻⁹

      P = 1.0 10⁵ W

b) intensity is defined as the ratio between power per area unit

      I = P / A

The area of a circle is

     A = π r²

     r = d / 2

     A = π (0.5 10⁻³)²

     A = 0.785 10⁻⁶ m²

     I = 1.0 10⁵ / 0.786 10⁻⁶

     I = 1.27 10¹¹ W / m²

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The voltage across an inductor ' L ' is

V = L · dI/dt .

I(t) = I(max) sin(ωt)

dI/dt = I(max) ω cos(ωt)

V = L · ω · I(max) cos(ωt)

L = 1.34 x 10⁻² H

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V = L · ω · I(max) cos(ωt)

V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)

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6 0
3 years ago
Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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