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1 year ago

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A piece of fine fiber with a diameter of =6.5 m is used to prop apart the edges of two perfectly flat 3.3-cm-long pieces of glas

s (see diagram). When the setup is illuminated from above with light of wavelength =590 nm , an interference pattern of alternating bright and dark bands will be seen in the reflected light. If the setup is viewed from high above, how many dark bands will be seen?

1 answer:

Anna71 [15]1 year ago

3 0

We need frequency:-

$\\ \rm\Rrightarrow \nu=\dfrac{c}{\lambda}$

$\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8m/s}{590\times 10^{-9}m}$

$\\ \rm\Rrightarrow \nu=0.0051\times 10^{17}Hz$

$\\ \rm\Rrightarrow \nu=5.1\times 10^{14}Hz$

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Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

T = $2\pi \sqrt{ \frac{L}{g} }$

T² = $4\pi ^2 \frac{L}{g}$ 4pi2 L / g

g = $4\pi ^2 \frac{L}{T^2}$

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

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g = $4\pi ^2 \frac{0.823}{1.823^2}$ 4 pi2 0.823 / 1.823 2

g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

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T = t / n

ΔT = $\frac{dT}{dt}$ Δt + $\frac{dT}{dn}$ ΔDn

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ΔT = Δt / n

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a more manageable way is with the relative error

$\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}$

we substitute

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Δg = 9.7766 ( $\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}$ )

Δg = 9.7766 (0.001215 + 0.0184)

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the absolute uncertainty must be true to a significant figure

Δg = 0.2 m / s2

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g ±Δg = (9.8 ± 0.2) m / s²

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Answer:

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(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

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$\Delta U=1.6\times10^{-19}\times639.2$

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$W=-8\times10^{-18}\ J$

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(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

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Answer:

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