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3 years ago

Using the data provided below, calculate the corrected wavelength for a spectroscope reading of 6.32.

Physics

1 answer:

weqwewe [10]3 years ago

6 0

Answer:

646.6 nm

Explanation:

Using the data given, we fit a linear model :

Spectroscope Readings Known Helium Wavelengths (nm)

3.65 388.9

4.75 486.6

4.90 501.6

5.65 587.6

6.45 667.8

7.00 706.5

Using technology like excel to fit the model, the regression model obtained is :

Y = 97.96582X + 27.48458

Where, y is the predicted or corrected wavelength value.

This mod could be used to calculate the corrected wavelength for a given Spectroscope value :

Given a Spectroscope value of 6.32 ; the corrected wavelength is obtained by replacing x in the equation by 6.32 and calculate y ;

Y = 97.96582X + 27.48458

x = 6.32

Y = 97.96582(6.32) + 27.48458

Y = 619.1439824 + 27.48458

Y = 646.6285624

Corrected wavelength value is : 646.6

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Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

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Where v is the velocity and r is the radius

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$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

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$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

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$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

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Then, equation 2 can be used

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$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

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Finally, equation 1 can be used:

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