Answer:
The higher the amplitude, the higher the energy. To summarise, waves carry energy. The amount of energy they carry is related to their frequency and their amplitude. The higher the frequency, the more energy, and the higher the amplitude, the more energy.
Answer:
Disruption to electricity power grid
Explanation:
We're looking a a solar flare. This will whip solar particles at high velocity into space and, If they are near earth, will interact with the earth's magnetic field. These magnetic changes will be measurable in the electric grid. Whether they are strong enough to cause "disruption" depends on a huge number of factors such as strength of and angles of the interacting magnetic fields and location of grid infrastructure,
Answer:
Doppler Theory
Explanation:
it's a theory regarding the change in wave frequency during the relative motion between a wave source and its observer.
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance 
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q =
= 
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
Answer:
Explanation:
is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:
A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:
is the centripetal force and is defined as:
Here 
is the proton's speed and 
is the radius of the circular motion. Replacing this in (1) and solving for r:
Recall that 1 J is equal to 
, so:
We can calculate from the kinetic energy of the proton:
Finally, we calculate the radius of the proton path:
