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1 year ago

Please help meeeeeeeeeeeeeeee

Physics

1 answer:

maxonik [38]1 year ago

5 0

probabilityAnswer:

2/27

Explanation:

The elk can not be eaten so we remove that from the probablity

so we have x/18

songbird = 4/18

mice = 6/18

4/18*6/18 = 2/27

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The Andromeda Galaxy is faintly visible to the naked eye in the constellation Andromeda. Suppose instead it were located in the

Kaylis [27]

Answer:

a) We could not see it at all.

Explanation:

The most distant object that can be seen is the andromeda galaxy, which we may have a slight view of. The andromeda galaxy is a large galaxy that along with the previous two is also part of the local group. Spiral-type galaxy that is approximately 250,000 light years in diameter (more than twice the diameter of the Milky Way!) And is about 2.9 million light years away from our galaxy. Because of its distance, we have difficulty visualizing this galaxy, we would have this difficulty even if the andromeda galaxy was in the center of the Milky Way, but maintaining its current distance. That is, even if the andromeda galaxy were located in the same direction in space as the center of the Milky Way (but still at its current distance), we could not see it at all.

8 0

3 years ago

Holes are being steadily injected into a region of n-type silicon (connected to other devices, the details of which are not impo

Tems11 [23]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $J_n = -0.864 \ A/cm^2$

Explanation:

Generally for an n-type semiconductor the current density is mathematically represented as

$J_n = -q * d_p * \frac{d p_{n_o}}{d_w}$

Here $p_{n_o}$ is mathematically represented as

$p_{n_o} = \frac{n_i^2}{n_d}$

=> $p_{n_o} = \frac{(1.5*10^{10})^2}{10^{16}}$

=> $p_{n_o} = 2.25 *10^{4} \ cm^{-3}$

$d p_{n_o} = P_n0 - p_{n_o}$

From the diagram $P_n0 = 10^8 * p_{n_o}$

=> $P_n0 = 10^8 * (2.25 *10^{4} )$

$d p_{n_o} = 10^8 * (2.25 *10^{4} ) - 2.25 *10^{4}$

$d p_{n_o} = 2.25 *10^{12} cm^{-3}$

So from $J_n = -q * d_p * \frac{d p_{n_o}}{d_w}$

substitute

$1.60 *10^{-19} \ C$ for q and $w_2 =50 nm = 50*10^{-9} m = 5*10^{-6} cm$

and from the diagram $w_1 =0 \ cm$

$J_n = -1.60 *10^{-19} *12 * \frac{2.25 *10^{12} }{ 5*10^{-6} - 0 }$

$J_n = -0.864 \ A/cm^2$

6 0

3 years ago

A student conducted an experiment to measure the acceleration due to gravity 'g' of a simple pendulum.

valina [46]

I think i used calulater and it gives me 47.5

5 0

3 years ago

When an electron falls from a higher to a lower energy level in an atom, the photon released has a wavelength of 121.6 nm. What

yarga [219]

Answer:

$\Delta E=1.64*10^{-18}J$

Explanation:

The energy difference between the energy levels involved in the transition of the electron is directly proportional to the frequency of the emitted photon:

$\Delta E=h\nu(1)$