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beks73 [17]
3 years ago
14

The allowed energies of a simple atom are 0.0 eV, 4.0 eV, and 6.0 eV. Part A What wavelength(s) appear(s) in the atom's emission

spectrum
Physics
1 answer:
Dmitriy789 [7]3 years ago
3 0

Answer:

3.1 × 10^- 7 m and 2.1 × 10^-7 m

Explanation:

First we must convert each value of energy to Joules by multiplying its value by 1.6 ×10^-19. After that, we can now obtain the wavelength from E= hc/λ

Where;

h= planks constant

c= speed of light

λ= wavelength of light

For 6.0ev;

E= 6.0 × 1.6 ×10^-19

E= 9.6 × 10^-19 J

From

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/9.6 × 10^-19

λ= 2.1 × 10^-7 m

For 4.0 eV

4.0 × 1.6 × 10^-19 = 6.4 × 10^-19 J

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/6.4 × 10^-19

λ= 3.1 × 10^- 7 m

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Glycerin at a temperature of 30 degrees celcius flows at a rate of 8×10−6m3/sthrougha horizontal tube with a 30mmdiameter. what
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The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>
  • We have the Poiseuille's formula,

                     Q=\frac{\pi r^4P}{8\beta l}

  • where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.
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                  P=3.284Pascals

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Learn more about the Poiseuille's formula here:

brainly.com/question/13180459

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3 0
1 year ago
What can happen if the president chooses to veto a law that has been passed? The law returns to being introduced in one of the h
andre [41]

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The president s decision can be overridden by a majority of Congress.

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4 0
3 years ago
A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

W = 3.12 J

Hope this helps!

4 0
3 years ago
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