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3 years ago

Question 1

Physics

1 answer:

romanna [79]3 years ago

3 0

Answer:

1.- the lithosphere, asthenosphere, and mesosphere.

2.- They both described the oceanic type.

3.- Crust, as its depth is less than 100 kilometers

4.- High pressure

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A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what angle relative to the normal

AleksAgata [21]

Answer:

a. 78 degree

Explanation:

According to Snell's Law, we have:

(ni)(Sin θi) = (nr)(Sin θr)

where,

ni = Refractive index of medium on which light is incident

ni = Refractive index of ethyl alcohol = 1.361

nr = Refractive index of medium from which light is refracted

nr = Refractive index of ethyl alcohol = 1.333

θi = Angle of Incidence

θr = Angle of refraction

So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:

θi = θc

when, θr = 90°

Therefore, Snell's Law becomes:

(1.361)(Sin θc) = (1.333)(Sin 90°)

Sin θc = 1.333/1.361

θc = Sin⁻¹ (0.9794)

θc = 78.35° = 78° (Approximately)

Therefore, correct answer will be:

a. 78 degree

8 0

3 years ago

DetermiOne of the lines in the Balmer series of the hydrogen atom emission spectrum is at 397 nm. It results from a transition f

larisa86 [58]

Answer:

a)n =7 , b) n = 5

Explanation:

The energy levels of the hydrogen atom is described by the Bohr model

$E_{n}$ = - 13.606 1/n²

This equation energy is given in elector volts and n is an integer

A transition occurs when the electro sees from a superior to a lower state

E₀ - $E_{n}$ = -13.606 (1/ $n_{f}$ ² - 1/n₀²)

Let's apply this expression

n₀ = 2

Let's look for the energy of the different levels and subtract it

n₀ $E_{n}$ (eV)

1 -13,606

2 -3.4015

3 - 1.5118

4 -0.850375

5 -0.54424

6 -0.3779

7 -0.2777

The wavelength of the transition is 397 nm = 397 10⁻⁹ m

The speed of light is related to wavelength and frequency

c = λ f

The Planck equation gives the energy of a transition

E = h f

E = h c /λ

Let's calculate

E = 6.63 10⁻³⁴ 3 10⁸/397 10⁻⁹

E = 5.01 10⁻¹⁹ J

Let's reduce to eV

E = 5.01 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

E = 3.1313 eV

Let's examine the possible transitions from the initial level ni = 2

ΔE = $E_{2}$ - $E_{n}$ = -3.1313

En = 3.13 -3.4015

$E_{n}$ = 0.2702 eV

When examining the table we see that the level that this energy has is the level of n = 7

Part B the transition is in the infrared

The frequency is 74 10¹² Hz

We use the Planck equation

E = h f

E = 6.63 10⁻³⁴ 74 10¹²

E = 4.9062 10⁻²⁰ J

E = 4.9062 10⁻²⁰ / 1.6 10⁻¹⁹

E = 0.3066 eV

We look for the level with the energy difference

ΔE = E₄- $E_{n}$ = 0.3066

$E_{n}$ = 0.3066 - 0.85037

$E_{n}$ = -0.54376 eV

When examining the table this energy has the level n = 5, therefore from this level the transition occurs

6 0

4 years ago

The distance of motion, or displacement, along a fault during an earthquake is known as:

k0ka [10]

A fault slip is the distance of motion, or displacement, along a fault during an earthquake. They can be classified by their relation to the horizontal, for example, if the fault slip happens primarily vertically it is a dip slip.

8 0

3 years ago

A car with an initial velocity of O m/s and a mass of 1500 kg reaches a velocity of 15 m/s in 5

Flura [38]

Answer:the answer can be find out by the following steps;

Explanation: