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Misha Larkins [42]
3 years ago
12

What do sound and light waves do when bouncing off objects

Physics
1 answer:
Paraphin [41]3 years ago
8 0
Radioactive waves is the answer
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The cardiovascular system moves blood through hour body. It is made up of the Heart pumping the blood that is circulated around the body through its networks of arteries, veins and capillaries. The cardiovascular and lymphatic systems together make up the circulatory system.
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A proton that has a mass m and is moving at +164 m/s undergoes a head-on elastic collision with a stationary carbon nucleus of m
Irina18 [472]
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:

m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,

m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂'  --> equation 1

The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is

(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2

Solving equations 1 and 2 simultaneously, v₁' =  -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
7 0
3 years ago
The Balmer series in hydrogen atom produces only infrared emission. O True O False
lorasvet [3.4K]

Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

n_{i}=2.

[tex]n_{f}=3,4,5,.....\infty.

Now the value of wavelength can be calculated as,

\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}.

Here, R=109677 cm^{-1}.

And n_{f}=3.

Now,

\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3}).

Therefore,

\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

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