5 m/s
30 divided by 6 is 5
Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:

In our problem we have
and
, so we can find the magnitude of the electric field:

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to

Therefore, the magnitude of the force acting on the proton will be

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
Answer:
high, low
Explanation:
- Energy always flows from a higher level to a lower level.
- It is analogous to the waterfall where waterfalls from a higher level to a lower level.
- So in the case of the pressure of the gas, when there are any numbers of molecules in a given volume of space. The gas is said to be at high pressure.
- When there are fewer molecules in the given volume. The gas is said to be at lower pressure.
- Due to a large number of atoms, the high-pressure gas exerts more force on the container than the force exerted by the low-pressure gas.
- If a hose is connected between these two containers, gas rushes from high pressure to the low pressure. Since the force exerted by the high-pressure gas is greater than that of low-pressure gas.
So, the wind tends to move from high-pressure areas to low pressure.
In the part of the spectrum our eyes can detect (a spectrum is an arry of entities, as light waves or particles, ordered in accordance with the magnitudes of a common physical property, as wavelength or mass) Hope this helps you :D