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Makovka662 [10]
2 years ago
5

Where is Earth’s magnetic south pole located? What about the magnetic north pole? What does this say about how a compass needle

will react to the poles?
Physics
1 answer:
Dominik [7]2 years ago
7 0

Answer:

Currently, the magnetic south pole lies about ten degrees distant from the geographic north pole, and sits in the Arctic Ocean north of Alaska. The north end on a compass therefore currently points roughly towards Alaska and not exactly towards geographic north.

Explanation:

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What is the energy of a photon whose frequency is 6.0 x 10^20?
omeli [17]

Answer:

3.75 MeV

Explanation:

The energy of the photon can be given in terms of frequency as:

E = h * f

Where h = Planck's constant

The frequency of the photon is 6 * 10^20 Hz.

The energy (in Joules) is:

E = 6.63 x10^(-34) * 6 * 10^(20)

E = 39.78 * 10^(-14) J = 3.978 * 10^(-13) J

We are given that:

1 eV = 1.06 * 10^(-19) Joules

This means that 1 Joule will be:

1 J = 1 / (1.06 * 10^(-19)

1 J = 9.434 * 10^(18) eV

=> 3.978 * 10^(-13) J = 3.978 * 10^(-13) * 9.434 * 10^(18) = 3.75 * 10^(6) eV

This is the same as 3.75 MeV.

The correct answer is not in the options, but the closest to it is option C.

6 0
3 years ago
Light propagate faster through medium “a” than medium “b”
dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1, n_2 are the index of refraction of the two mediums

\theta_1, \theta_2 are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)

Here we want the direction of propagation of the light ray not to change: this means that it must be

sin \theta_1 = sin \theta_2 (1)

However, here we have two mediums "a" and "b" with different index of refraction, so

n_1\neq n_2

Therefore the only angle that can satisfy eq.(1) is

\theta_1 = \theta_2 = 0

So, the light must hit the surface perpendicular to the interface between the two mediums.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

3 0
3 years ago
What’s the equation for this
aleksley [76]
Where is the question
8 0
3 years ago
Substances released into the air are known as
irakobra [83]

Your answer is Emissions

3 0
3 years ago
Read 2 more answers
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
3 years ago
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