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user100 [1]
3 years ago
15

Extra Practice In Exercises 1-4, use the diagram. 1. Give two other names for CD. ​

Physics
1 answer:
lana [24]3 years ago
6 0

Answer: I think it’s point cd and line cd

Explanation:

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2. Summarize how the force of gravity
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In summary, gravity is the force that creates the pressure to fuse atoms, which makes the stars shine. Eventually the temperature is high enough that the star starts fusing hydrogen into helium. When the outward pressure produced by the heating of the gas by fusion energy balances gravity, a stable star is formed.
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Which of the following numbers most closely represents the number of galaxies in the universe?
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There are an estimated 100 billion, so the answer would be D.) Billions
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3 years ago
A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.2
julsineya [31]

Answer:

The wavelength of next line in the series will be 397.05 nm

Explanation:

From Rydberg equation;

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

where;

λ is the wavelength

n lines in the series

RH is Rydberg constant = 1.097 x 10⁷ m⁻¹

Also at a given maximum wavelength, we can determine the first line n₁ in the series

\frac{1}{\lambda_{max}R_H} = \frac{1}{n_1^2} -\frac{1}{(n_1 +1)^2} \\\\

\frac{1}{\lambda_{max}R_H} = \frac{2n_1+1}{n_1^2(n_1 +1)^2} \\\\\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}

Given;

maximum wavelength = 656.46 nm

\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\656.46 *10^{-9}*1.097*10^7 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\7.2 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}

Now, test for different values of n that will be equal to 7.2

let n₁ = 1

\frac{n_1^2(n_1 +1)^2}{2n_1+1} =  \frac{(1)^2(1 +1)^2}{2(1)+1} = 1.3\\

n₁(1) ≠ 7.2

Again, let n₁ = 2

\frac{n_1^2(n_1 +1)^2}{2n_1+1} =  \frac{(2)^2(2 +1)^2}{2(2)+1} = 7.2\\

∴ n₁(2) = 7.2

For the least wavelength given as 410.29 nm, n = ?

\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{410.29*10^{-9}} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{n_2^2})\\\\\frac{1}{4.5} =\frac{1}{4} -\frac{1}{n_2^2}\\\\\frac{1}{n_2^2} =\frac{1}{4}  -\frac{1}{4.5} \\\\\frac{1}{n_2^2}  = 0.0277778\\\\n_2^2 = \frac{1}{0.0277778} \\\\n_2^2 = 36\\\\n_2= \sqrt{36}\  = 6

next line in the series will be 7

The wavelength of next line in the series will be;

\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{7^2})\\\\\frac{1}{\lambda}  =  1.097*10^7(0.22959)\\\\\frac{1}{\lambda}   = 2518602.3\\\\\lambda = 397.05 \ nm

Therefore, the wavelength of next line in the series will be 397.05 nm

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3 years ago
What tells you the strength of a wave?​
castortr0y [4]
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all these factors determines the strength of a wave.

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3 0
3 years ago
In a photoelectric effect experiment, it is observed that violet light does not eject electrons from a particular metal. Next, r
liubo4ka [24]

Answer:

No ejection of photo electron takes place.

Explanation:

When a photon of suitable energy falls on cathode, then the photoelectrons is emitted from the cathode. This phenomenon is called photo electric effect.

The minimum energy required to just  eject an electron is called work function.

The photo electric equation is

E = W + KE

where, E is the incident energy, W is the work function and KE is the kinetic energy.

W = h f

where. h is the Plank's constant and f is the threshold frequency.

Now, when the violet light is falling, no electrons is ejected. When the red light is falling, whose frequency is less than the violet light, then again no photo electron is ejected from the metal surface.

6 0
3 years ago
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