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KIM [24]
3 years ago
13

When an electron in a certain excited energy level in a one-dimensional box of length 2.00 Å makes a transition to the ground st

ate, a photon of wavelength 8.79 nm is emitted. Find the quantum number of the initial state?​
Physics
1 answer:
Fiesta28 [93]3 years ago
5 0

Answer:

Calculate the wavelength associated with an electron with energy 2000 eV.

Sol: E = 2000 eV = 2000 × 1.6 × 10–19 J

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The velocity of the transverse waves produced by an earthquake is 5.08 km/s, while that of the longitudinal waves is 8.3312 km/s
ASHA 777 [7]

Answer:

727.67 km

Explanation:

Sine they have Same distance D

distance = speed * time

D = 5.08t

D = 8.3312(t+55.9)

so

5.08t = 8.3312(t+55.9) t in

3.2512t = 465.71

t = 143.2s

Subtitute t

D=5.08 t

= 5.08 × 143.2

= 727.67km

7 0
3 years ago
As beverages are produced what action will increase the solubility of co2 in them most?
Alla [95]

Answer: D. decreasing the temperature

Explanation:

8 0
4 years ago
Read 2 more answers
A long electric cable is suspended above the earth and carries a current of 345 A parallel to the surface of the earth. The eart
jok3333 [9.3K]

Answer:

0.906 N

Explanation:

Formula for magnetic force acting on current carrying cable:

F = IBLsin(\theta)

Where I = 345A is the current in the wire, B = 5.6*10^{-5} T is the magnetic magnitude generated by Earth. L = 46.9 m is the cable length. \theta = 88.2^o is the angle between vector B and cable direction.

F = 345*5.6*10^{-5}*46.9*sin(88.2^o)

F = 0.906 N

7 0
3 years ago
A rogue wave is a random monstrous wave that occurs unexpectedly in the ocean.
Tpy6a [65]

A wave with a large amplitude

a wave check

8 0
3 years ago
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Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.4 cmcm and a current of 12 AA. The bigger
Free_Kalibri [48]

Answer:

Explanation:

Given that,

Current in loops are

i1 = 12A

i2 = 20A

The loops are 3.4cm apart

The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop

Magnetic Field is given as

B= μoi/2πr

Where,

μo is a constant = 4π×10^-7 Tm/A

r is the distance between the two wires

i is the current in the wires

B is the magnetic field

NOTE

Field due to large loop should be equal to the smaller loop.

B1 = B2

μo•i1 / 2π•r1 = μo•i2 / 2π•r2

Then, μo, 2π cancels out, so we have

i1 / r1 = i2 / r2

Make r2 subject of formula

i1•r2 = i2•r1

r2 = i2•r1 / i2

r2 = 20×3.4/12

r2 = 5.67cm

The radius of the bigger loop is 5.67cm.

4 0
3 years ago
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