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Vera_Pavlovna [14]
1 year ago
14

An 11.8 eV photon is absorbed by an electron in an atom, causing the electron to be excited from level 1 to level 3. The electro

n then emits a 4.1 eV photon and shifts to level 2. What photon will be emitted when the electron moves back to level 1?
a. 15.9 eV
b. 11.8 eV
c. 7.7 eV
d. 4.1 eV
Physics
1 answer:
densk [106]1 year ago
6 0

Answer:

c. 7.7 eV

Explanation:

The computation of the photon emitted is shown below:

Given that

From level 1 to level 3, the energy absorbed by electrons is 11.8eV

And, when from level 3 to level3, the energy released by electron is 4.1eV

So, from level 2 to level 1, the energy released is

= 11.8 - 4.1

= 7.7

= 7.7eV

Hence, the third option is correct

The same is to be considered

And, the other options are wrong

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When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec
bixtya [17]

Answer:

1340.2MW

Explanation:

Hi!

To solve this problem follow the steps below!

1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid

W=αhQ

α=specific weight for water =9.81KN/m^3

h=height=220m

Q=flow=690m^3/s

W=(690)(220)(9.81)=1489158Kw=1489.16MW

2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator

Wr=(0.9)(1489.16MW)=1340.2MW

the maximum possible electric power output is 1340.2MW

3 0
2 years ago
A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and
scoray [572]

Answer:

F_{allow} = 208.15kN

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

K_{allow} =\frac{K_c}{N}

We know that K_c is 33Mpa*m^{0.5} and our Safety factor is 2,

K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

\alpha = 0.5*L_c = 0.5*4mm = 2mm

<em>For the panel</em>

\gamma = 0.5*W = 0.5*250mm = 125mm

To find now the goemetry factor we need to use this equation

\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1

That allow us to determine the allowable nominal stress,

\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}

\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

F_{allow}=\sigma_{allow}*L_c*W

F_{allow} = 208.15*250*4

F_{allow} = 208.15kN

That is the maximum tensile load before a catastrophic failure.

4 0
2 years ago
A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitca
Hatshy [7]

Answer:

0.0367

Explanation:

The loss in kinetic energy results into work done by friction.

Since kinetic energy is given by

KE=0.5mv^{2}

Work done by friction is given as

W= umgd

Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.

Making u the subject of the formula then we deduce that

u=\frac {v^{2}}{2gd}

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367

Therefore, the coefficient of kinetic friction is approximately 0.0367

7 0
2 years ago
Seven seconds after a brilliant flash of lightning, thunder shakes the house. approximately how far was the lightning strike fro
tangare [24]
Very roughly 7,700 feet ... about 1.5 miles.
8 0
2 years ago
A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

T + mg =\frac{mv^{2}}{r}

Inserting the values

T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}

T = 31.1 N

7 0
2 years ago
Read 2 more answers
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