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pantera1 [17]
3 years ago
5

The components of a 15 meter per second velocity at an angle of 60 degres above the horizontal

Physics
1 answer:
Neko [114]3 years ago
6 0
X= 7.5 m/s
Y= 12.99 m/s 
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When drawing a Bohr model for an element that has 16 electrons, how many electrons would be placed in the third energy level?
lubasha [3.4K]
There would be 6 electrons placed on the third energy level.
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3 years ago
The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible
Leto [7]

Answer:

a)   298.5 nm ,  522.4 nm  and b)  radiation frequency does not change

Explanation:

When electromagnetic radiation reaches a medium with a different index of refraction, the medium vibrates the molecules, as if it were a resonance process, whereby the medium vibrates at the same frequency as the incident light.

On the other hand, when the light reaches another medium its average speed within the medium changes, it is now less than the speed of light in a vacuum (c) for this to happen as we saw that the frequency is constant there must be a change in the wavelength of the radiation that is characterized by the ratio

    λₙ = λ₀ / n

    λₙ = 400 nm    in the void

    λₙ = 400 / 1.34

    λₙ= 298.5 nm

   λ₀ = 700 nm

   λₙ = 700 / 1.34

   λₙ = 522.4 nm

The radiation frequency does not change

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3 years ago
Plz help i'm desperate! Why are all people on Earth members of the same biological Species?
Alex Ar [27]

Answer:

D.) All people can mate with one another and have children

Explanation:

7 0
3 years ago
A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec
kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, \lambda=\dfrac{\lambda_ox_o}{x}

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

dE=\dfrac{k\ dq}{x^2}..........(1)

The linear charge density is given by :

\lambda=\dfrac{dq}{dx}

dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx

Integrating equation (1) from x = x₀ to x = infinity

E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx

E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}

E=\dfrac{k\lambda_o}{2x_o}

Hence, this is the required solution.

5 0
3 years ago
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What is 0.94kg divided by 2.4n
LuckyWell [14K]

Answer:

0.39166666666

Explanation:

7 0
3 years ago
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