The frequency of a wave becomes higher due to the object moving at a fast pace coming towards you with shorter wavelengths (depending on the speed) aka the Doppler Effect.
Hope this helps
The average electric current in the lightning will be 8 × 
A
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Why Lightning Conductors on top of a tower ?</h3>
The lightning conductors are long metal strips running from the spike end of a conductor on the top of a building to the earth. They are used to prevent buildings from destruction when struck by thunder or lightning.
Given that a lightning strike can transfer as much as electrons from the cloud to the ground. if the strike lasts 2ms, to calculate the average electric current in the lightning, we will first consider the charge released.
one charge = 1.6 × 
C
Average current I = Q/t
Where
- Q = charge = 1.6 × C
- t = time = 2ms = 2 ×
s
Substitute all the parameters into the formula
I = 1.6 × C ÷ 2 ×
I = 8 × A
Therefore, the average electric current in the lightning will be 8 × A
Learn more about Lightning here: brainly.com/question/3183045
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Explanation:
Optics is a branch of physics that is the study of light and vision. ... The branch of physics dealing with the nature and properties of electromagnetic energy in the light spectrum and the phenomena of vision. In the broadest sense, optics deals with infrared light, visible light, and ultraviolet light.
A probability contour represents a bounded, finite volume about a nucleus in which there is a substantial probability of finding the electron is TRUE.
The rectangular of the wave characteristic, ψ2 , represents the opportunity of locating an electron in a given vicinity within the atom. probability finding electron at nucleus is zero.
An orbital is a mathematical feature that has a fee at all points in space. The magnitude squared of that fee is the chance according to volume of locating an electron in that volume of area.
The probability of locating electron in nodes is 0, but there are electron densities on all round nodes. The node is a point or a floor (relying at the type of node) so the volume of the vicinity wherein ψ=zero is zero. we want to position V=zero and we get P=zero so the chance of finding the electron on the node is zero.
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Answer:
Given: Vi = 382 km/h, Vf = 0 km/h, Mc = 705 kg, Md = 65 kg, Δt = 12
Required: Δx
F = Δp / Δt
= ![\frac{(Mc+Md)Vf-(Mc+Md)Vi}{t} \\\\= 6.81 * 10x^{3} N [left]\\\\x=\frac{1}{2} (Vi+Vf)\\ \\ = 637m[right]](https://tex.z-dn.net/?f=%5Cfrac%7B%28Mc%2BMd%29Vf-%28Mc%2BMd%29Vi%7D%7Bt%7D%20%5C%5C%5C%5C%3D%206.81%20%2A%2010x%5E%7B3%7D%20N%20%5Bleft%5D%5C%5C%5C%5Cx%3D%5Cfrac%7B1%7D%7B2%7D%20%28Vi%2BVf%29%5C%5C%20%5C%5C%20%3D%20637m%5Bright%5D)
