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Aneli [31]
3 years ago
10

What is the density of an object if it has a volume of 162 ml with a mass(273) that is shown in the picture below?

Physics
2 answers:
creativ13 [48]3 years ago
7 0

Answer:

1.685185185185185 (you can round to whatever your teacher asks)

Explanation:

Formula for Density

D=m/v (Density equals mass over volume)

273/162

(/) means division.

e-lub [12.9K]3 years ago
7 0
It’s 1.65 I just rounded it real quick but that the short answer
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You discover a new type of microscopic, atom-like object. The energy levels for this object are given by En = E1 n , where E1 =
Studentka2010 [4]

Answer:

a) Excitation energy for the third excited state = -112 eV

b) The amount of energy required to cause an object in the third excited state to become unbound = 112 eV

c) Max number of photons emitted as the object de - excites from the third excited state to the ground state = 6

d) Maximum wavelength photons = 532.286 nm

Minimum wavelength photons = 59.14 nm

Explanation:

From the question, the energy levels are given by E_{n} = E_{1} n

Where the energy level for the ground state is given by E_{1} = -28.0 eV

a) Excitation energy for the third excited state, n = 4

E_{4} = 4 * E_{1}

E_{4} = -28 * 4

E_{4} = -112 eV

b) The amount of energy required to cause an object in the third excited state to become unbound

E_{4} = 112 eV

c) Max number of photons emitted = \frac{n(n-1)}{2}

Max number of photons emitted  = \frac{4(4-1)}{2}

Max number of photons emitted  = 2 * 3 = 6

d) \frac{1}{\lambda_{max} } = \frac{28}{hc} [\frac{1}{3} -\frac{1}{4}]

hc = 1242 eV-nm

\frac{1}{\lambda_{max} } = \frac{28}{1242} [\frac{1}{3} -\frac{1}{4}]

\lambda_{max} = 532.286 nm

Maximum wavelength photons = 532.286 nm

\frac{1}{\lambda_{min} } = \frac{28}{hc} [\frac{1}{1} -\frac{1}{4}]

\frac{1}{\lambda_{min} } = \frac{28}{1242} [\frac{1}{1} -\frac{1}{4}]

\lambda_{min} = 59.14 nm

Minimum wavelength photons = 59.14 nm

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3 years ago
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