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3 years ago

What lasts longer and why?: A solar eclipse or a lunar eclipse?

1 answer:

5 0

Lunar eclipse can lasts longer because it lasts around 3 hours but can last as long as four hours. Solar eclipse last shorter than lunar eclipse because the moon only creates a small shadow on the earth and the moon travels through the suns vision (as seen on earth) in a short amount of time.

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What is the ground-state electron configuration of a neutral atom of titanium?

Georgia [21]

Answer:

$\bold {1s^22s^22p^63s^23p^64s^23d^2}$

Explanation:

electronic configuration of an atom is the spatial arrangement of the electrons around the nucleus in the energy orbits

We know that the atomic number of Titanium is 22

The electronic configuration of Titanium is

$1s^22s^22p^63s^23p^64s^23d^2$

3 0

3 years ago

On a velocity vs. time graph, what does it mean when the line crosses over the x-axis?

Phantasy [73]

On a velocity - time graph, if the line crosses the x - axis it depicts that the object has started moving in the opposite direction.

8 0

3 years ago

Which is a charcharacteristic of all ions?

ankoles [38]

An ion is created by the transfer of electrons. The metals give away the elections and become positively charged. The non - metals take on electrons.

Balance.

So an ion is any atom that either gives away or takes on electrons.

8 0

3 years ago

Use Snell's Law to solve the following:

azamat

Answer:

1.171

Explanation:

if n₁sinΘ₁=n₂sinΘ₂, then n₂=n₁sinΘ₁ / sinΘ₂;

$n_2=\frac{1.5*sin45}{sin65}=\frac{1.5*0.707}{0.906} =1.1705$

3 0

3 years ago

What is the wavelength, in nm, of the line in the hydrogen spectrum when one n value is 3 and the other n value is 6?

iren [92.7K]

Answer:

$\lambda=1090nm$

Explanation:

Rydberg formula is used to calculate the wavelengths of the spectral lines of many chemical elements. For the hydrogen, is defined as:

$\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where $R_H$ is the Rydberg constant for hydrogen and $n_1$ , $n_2$ are the lower energy state and the higher energy state, respectively.

$\frac{1}{\lambda}=1.10*10^{7}m^{-1}(\frac{1}{3^2}-\frac{1}{6^2})\\\frac{1}{\lambda}=9.17*10^{5}m^{-1}\\\lambda=\frac{1}{1.09*10^{6}m^{-1}}\\\lambda=1.09*10^{-6}m*\frac{10^{9}nm}{1m}\\\lambda=1090nm$

4 0

3 years ago