Answer:
3,4,5,6,7,8,9,10
3. lithium
4. beryllium
5. boron
6. carbon
7. nitrogen
8. oxygen
9. fluorine
10. neon
those are the numbers in period 2 on the periodic table.
Explanation:
Answer:
Explanation:
The strengthcompassion field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines. The direction of the electric field is tangent to the field line at any point in space. Field lines can never cross. These pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would accelerate if placed upon the line. As such, the lines are directed away from positively charged source charges and toward negatively charged source charges.
Rules for drawing electric field lines
1. Electric field lines are always drawn from High potential to
low potential.
2. Two electric field lines can never intersect each other.
3. The net electric field inside a Conductor is Zero.
4. Electric field line from a positive charge is drawn radially outwards and from a negative charge radially inwards.
5. The density of electric field lines tells the strength of the electric field at that region.
6. Electric field lines terminate Perpendicularly to the surface of a conductor.
A vector quantity has a direction and a magnitude, while a scalar has only a magnitude. You can tell if a quantity is a vector by whether or not it has a direction associated with it.
So, electric fields are vector quantity due to the fact any student can tell you that a compass is used to determine which direction is north.
Since the compass always point northward, then it has a direction and magnitude and so it is a vector quantity
Answer:
true
Explanation:
i believe that's the answer hope that helps
Answer:
Electric field, E = 40608.75 N/C
Explanation:
It is given that,
Mass of electrons, 
Initial speed of electron, u = 0
Final speed of electrons, 
Distance traveled, s = 6.3 cm = 0.063 m
Firstly, we will find the acceleration of the electron using third equation of motion as :
Now we will find the electric field required in the tube as :
E = 40608.75 N/C
So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.
