Answer:
The electromagnetic wave that travels the fastest through space is gamma ray
Explanation:
Electromagnetic wave is a type of wave which does not require material medium for its propagation. Examples of electromagnetic waves according to increasing frequency of the waves are gamma ray, x-ray, ultra violet ray, infra red, visible light, micro wave and radio waves.
Frequency of a wave is inversely proportional to period of oscillation of the wave. The higher the frequency of a wave, the shorter the period of oscillation. Gamma ray has the highest wave frequency in electromagnetic spectrum and shorter period of oscillation, thereby causing it to have the highest penetration power.
Answer:
The reaction force is the force exerted by your body on the earth pulling upwards.
Explanation:
This can be explained using Newton's third law of motion which states that if object A exerts a force on object B, object B exerts a force of equal magnitude and opposite direction on object A. One characteristics of these force are that they act o 2 different bodies. In this example, the earth exerts a force on you. The Newton's reaction force would be you exerting a force of equal magnitude on the earth.
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Explanation:
Given that,
Mass of a body, m = 1 kg
Force constant, k = 16 N/m
We need to find the angular frequency and the frequency of oscillation.
(a) The angular frequency of a body is given by :
(b) The frequency of oscillation is given by :
Hence, this is the required solution.
Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by
where 
Difference in potential between the points is
![kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}](https://tex.z-dn.net/?f=kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7B0.2%5Ctext%7B%20m%7D%7D%20-%5Cleft%28%20-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D%20%3D%20%5Cdfrac%7BkQ%7D%7B0.2%5Ctext%7B%20m%7D%7D%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E9%5Ctext%7B%20F%2Fm%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%5Ctext%7B%20C%7D%7D%7B0.2%5Ctext%7B%20m%7D%7D)
(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
![270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]](https://tex.z-dn.net/?f=270%5Ctimes10%5E3%20%3D%20kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7Bx%7D-%5Cleft%28-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D)
The charge chould be moved to infinity
