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navik [9.2K]
3 years ago
15

how do the retinas of the eyes of night hunting animals differ from the retinas of animals that hunt during the day time

Physics
2 answers:
TEA [102]3 years ago
5 0
Hello,

Night-hunting animals have more rod cells than day-hunting animals. This is most likely because night-hunting animals have to be able to see better than day-hunting animals.
maksim [4K]3 years ago
4 0
Night hunting animals have more rod cells in their retinas which allows them to see better in the dark.
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A very long, solid insulating cylinder has radius R; bored along its entire length is a cylindrical hole with radius a. The axis
lawyer [7]

Answer:

The value of the electric field is E_{net} = \dfrac{r \textbf{b}}{2\epsilon_{0}}

Explanation:

We know that the electric field inside a solid cylinder at a distance \textbf{r} from the centre is given by

E = \dfrac{\rho \textbf{r}}{2 \epsilon_{0}}

Let's consider the cross-section of the cylinder as shown in the figure. Let `O' be the centre of the long solid insulating cylinder having radius 'R'. Also consider that O' be the cetre of the hole of radius 'a' situated at a distance 'b' from 'O'. Given, the volume charge density of the material is 'r'. So, the volume charge density inside the hole will be '-r'. Let's consider 'P' be any arbitrary point inside the hole situated at a distance 's' from O'.

So, the electric field 'E_{O}' due to the long cylinder at point 'P' is given by

E_{O} = \dfrac{r \textbf{c}}{2 \epsilon_{0}}

and the electric field 'E_{O'}'due to the hole at point 'P' is given by

E_{O'} = \dfrac{\rho \textbf{s}}{2 \epsilon_{0}}

So the net electric field (E_{net}) inside the hole is given by

E_{net} = E_{O} - E_{O'} = \dfrac{r}{2\epsilon_{0}}(\textbf{c - s}) = \dfrac{r \textbf{b}}{2\epsilon_{0}}

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4 years ago
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salantis [7]

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Explanation :

The formula used for orbital angular momentum is:

L=\sqrt{l(l+1)}\hbar

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L = orbital angular momentum

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As we are given the electronic configuration of Fe is, [Ar]3d^64s^2

Its most energetic electron will be for 3d electrons.

The value of azimuthal quantum number(l) of d orbital is, 2

That means, l = 2

Now put all the given values in the above formula, we get:

L=\sqrt{2(2+1)}\hbar

L=\sqrt{6}\hbar

Therefore, the magnitude of the orbital angular momentum for its most energetic electron is, \sqrt{6}\hbar

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3 years ago
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