Answer:
V0 = 44.97m/s
Height above fence = 66.67m
Explanation:
The detailed calculation is done as shown in the attachment
Since carbon has 4 valence electrons, a total of 4 bonds will be drawn withing methane, one between each H and the center C
Hope this helps
Answer:
Amplitude and Frequency
Explanation:
Analog signals are composed of continuous waves that can have any values for frequency and amplitude. These waves are smooth and curved.
Radio transmissions are a combination of two kinds of waves: audio frequency waves that represent the sounds being transmitted and radio frequency waves that "carry" the audio information. All waves have a wavelength, an amplitude and a frequency as shown in the figure. These properties of the wave allow it to be modified to carry sound information.
The two most common types of modulation used in radio are amplitude modulation (AM) and frequency modulation (FM). Frequency modulation minimizes noise and provides greater fidelity than amplitude modulation, which is the older method of broadcasting . Both AM and FM are analog transmission systems, that is, they process sounds into continuously varying patterns of electrical signals which resemble sound waves.
Answer:
or 0.32 μm.
Explanation:
Given:
The radiations are UV radiation.
The frequency of the radiations absorbed (f) = 
The wavelength of the radiations absorbed (λ) = ?
We know that, the speed of ultraviolet radiations is same as speed of light.
So, speed of UV radiation (v) = 
Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

Now, expressing the above equation in terms of wavelength 'λ', we have:

Now, plug in the given values and solve for 'λ'. This gives,
![\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B3%5Ctimes%2010%5E8%5C%20m%2Fs%7D%7B9.38%5Ctimes%2010%5E%7B14%7D%5C%20Hz%7D%5C%5C%5C%5C%5Clambda%3D3.2%5Ctimes%2010%5E%7B-7%7D%5C%20m%5C%5C%5C%5C%5Clambda%3D3.2%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2010%5E%7B6%7D%5C%20%5Cmu%20m%5C%20%5B1%5C%20m%3D10%5E6%5C%20%5Cmu%20m%5D%5C%5C%5C%5C%5Clambda%3D3.2%5Ctimes%2010%5E%7B-1%7D%3D0.32%5C%20%5Cmu%20m)
Therefore, the wavelength of the radiations absorbed by the ozone is nearly
or 0.32 μm.