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True [87]
2 years ago
5

Which electromagnetic has the shortest wavelength?.

Physics
1 answer:
Marina CMI [18]2 years ago
7 0

Answer:

Gamma rays

Explanation:

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Gemiola [76]

Answer:a computer , machine forcery,0,push

Explanation:

8 0
2 years ago
Positive feedback interactions in earth’s systems are always a result of human action. T/F
VLD [36.1K]

Positive feedback interactions in earth’s systems are always a result of human action is a FALSE statement.

<u>Explanation:</u>

Earth is a unstable equilibrium which tends to move out of equilibrium, but several other factors try to bring it back in equilibrium again. Earth has a different actions going on both on its surface and also inside it.

Human can alter, or can modify a very small part of the events that occur on earth’s surface. But they don’t have any control on what’s going inside earth’s core. So positive feedback interactions are not only the results of human interaction, but also different other factors.  

4 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
Calculate the wavelengths of the first five emission lines of the Balmer series for hydrogen
In-s [12.5K]

Answer:

Explanation:

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe. In quantum physics, when electrons transition between different energy levels around the atom (described by the principal quantum number, n) they either release or absorb a photon. The Balmer series describes the transitions from higher energy levels to the second energy level and the wavelengths of the emitted photons. You can calculate this using the Rydberg formula.

5 0
3 years ago
What happens to ocean water as it moves from Antarctica to the equator?
creativ13 [48]
It becomes less dense and rises to the surface

7 0
3 years ago
Read 2 more answers
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