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bagirrra123 [75]

3 years ago

10

What is frequency of light is emitted when an electron jumps into the smallest orbit of hydrogen, coming from a very large radiu

s (assume infinity)?

1 answer:

mrs_skeptik [129]3 years ago

7 0

Answer:

$3.28403\times 10^{15}\ Hz$

Explanation:

$n_1$ = 1

$n_2=\infty$

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

The energy is given by

$E(n)=E_1(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})\\\Rightarrow E(1)=13.6\ eV$

Energy is also given by

$E=hf\\\Rightarrow f=\dfrac{E}{h}\\\Rightarrow f=\dfrac{13.6\times 1.6\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=3.28403\times 10^{15}\ Hz$

The frequency of light emitted would be $3.28403\times 10^{15}\ Hz$

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What is the static friction force??

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2 years ago

A beach ball moving with the speed of 1.27 m/s rows of a pier and hits the water 0.75 m from the end the pier how high is the

Vera_Pavlovna [14]

Answer:

75.8

Explanation:

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3 years ago

After the big bang, atoms in gas clouds experienced a greater gravitational pull to each other than atoms in other regions of th

allsm [11]

Answer:
These are the two statements with scientific facts that explain the described phenomenon
<span>
Gravitation between two objects increases when the distance between them decreases.</span>

When the mass of an object increases, its gravitational pull also increases.

Justification:

Those two facts are represented in the Universal Law of Gravity discovered by the scientific Sir Isaac Newton (1642 to 1727) and published in his book <span>Philosophiae naturalis principia mathematica.</span>

That law is represented by the equation:

F = G × m₁ × m₂ / d²

The product of the two masses on the numerator accounts for the fact that the gravitational force is directly proportional to the product of the masses, which is that as the masses increase the attraction also increase.

The term d² (square of the distance that separates the objects) in the denominator accounts for the fact that the gravitational force is inversely proportional to the square of the distance; that is as the separation of the objects increase the gravitational force decrease.

6 0

3 years ago

4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=mV_{o}+MU_{o}$ (2)

$p_{f}=(m+M)V_{f}$ (3)

$m=1200 kg$ is the mass of the first car

$V_{o}=20 m/s$ is the velocity of the first car, to the North

$M=1400 kg$ is the mass of the second car

$U_{o}=-22 m/s$ is the mass of the second car, to the South

$V_{f}$ is the final velocity of both cars after the collision

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}$ (6)

Finally:

$V_{f}=-2.61 m/s$ (7) This is the resulting velocity of the wreckage, to the south

7 0

3 years ago