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podryga [215]
3 years ago
7

If the width of the Moon in the sky is 32 arc-minute, and the width of the Sun in the sky is 30 arc-minute, what kind of solar e

clipse would an earth-based observer witness, assuming the observer is perfectly aligned with the Sun and Moon?
Physics
1 answer:
shusha [124]3 years ago
8 0

Answer:

Total Solar Eclipse      

Explanation:

Solar eclipse occurs when moon covers the disc of the Sun. Annular eclipse occurs when the width of the moon in the sky is smaller than the width of the Sun. When the width of the moon is larger, the moon would cover the Sun completely and Total solar eclipse is said to occur.

The width of the Moon in the sky is 32 arc-minute and that of the Sun is 30 arc-minute. The Moon would fully cover the Sun because the width of the Moon is larger than the Sun. Thus, Total Solar eclipse would occur.

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After watching this video, Blake, a student in an introductory physics class, makes the following claim: The acceleration and ve
Maru [420]

Answer:

Please see below as the answer is self- explanatory.

Explanation:

  • Any time that an object changes direction  (from leftward to rightwward, or from upward to downward) the velocity must be zero just for one instant, when is on the verge of changing the direction.
  • This  is needed because velocity changes as a continuous function of time, so it needs to cross the t-axis when passing from positive to negative or vice versa.
  • However, the claim that in the moment that velocity is zero, the acceleration is also zero, is false.
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3 years ago
Which of the following is an example of sustainable use of natural resources? Question 21 options: 1) clear-cutting a forest 2)
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goldfiish [28.3K]

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8 0
3 years ago
The spectrum from a hydrogen vapour lamp is measured and four lines in the visible light range are observed. These lines are the
Elan Coil [88]

Answer:

10942249.24 m^{-1}

Explanation:

Rydberg's formula is used to describe the wavelengths of the spectral lines of chemical elements similar to hydrogen, that is, with only one electron being affected by the effective nuclear charge. In this formula we can find the rydberg constant, knowing the wavelength emitted in the transcision between two energy states, we can have a value of the constant.

\frac{1}{\lambda}=Z^2R(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})

Where \lambda it is the wavelength of the light emitted, R is the Rydberg constant, Z is the atomic number  of the element and n_{1} n_{2} are the states where n_{1}.

In this case we have Z=1 for hydrogen, solving for R:

R=\frac{1}{\lambda}*(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})^{-1}\\R=\frac{1}{658.9*10^{-9}m}*(\frac{1}{2^2}-\frac{1}{3^2}})^{-1}\\R=1.52*10^6m^{-1}*(\frac{36}{5})=1.09*10^7 m^{-1}=10942249.24m^{-1}

This value is quite close to the theoretical value of the constant R=10967758.34 m^{-1}

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