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2 years ago

How did bohr propose light was emitted from a material?

Physics

1 answer:

Artemon [7]2 years ago

6 0

Answer:

when an electron made a transition from an outer orbit to one closer to the nucleus

Explanation:

Bohr amended that view of the motion of the planetary electrons to bring the model in line with the regular patterns (spectral series) of light emitted by real hydrogen atoms. ... Light, he proposed, radiated from hydrogen atoms only

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A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli

fredd [130]

Answer:

Part a)

$A = 1.78 m$

Part b)

$f = 2 rev/s$

Explanation:

Part A)

As we know that time period of the motion is given as

$T = 2.68 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.68}$

$\omega = 2.34 rad/s$

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

$mg = m\omega^2 A$

so we have

$A = \frac{g}{\omega^2}$

$A = \frac{9.81}{2.34^2}$

$A = 1.78 m$

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

$mg = m\omega^2 A$

$g = \omega^2 (0.0623)$

$\omega = 12.5 rad/s$

so now we have

$2\pi f = 12.5$

$f = 2 rev/s$

3 0

2 years ago

Which choice has the least thermal energy?

natulia [17]

It’s A lolololololol

7 0

3 years ago

Read 2 more answers

A cannon is fired with an initial horizontal velocity of 20 m/s and initial vertical velocity of 25 m/s. After 3s in the air, th

Vladimir [108]

Answer:

60 m

Explanation:

After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.

The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.

7 0

2 years ago

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iragen [17]

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8 0

2 years ago

Read 2 more answers

Work and Power Practice Calculations

Elza [17]

Answer:

1. W = F d = 20 N * 6 m = 120 J

2. F = W / d = 60 J / 2 m = 30 N

3. d = W / F = 350 J / 85 N = 4.12 m

4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts

5. W = P t = 500 W * 120 sec = 60,000 J

6. t = W / P = 550 J / 310 W = 1.77 sec

5 0

1 year ago