The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.
Light because it is an electromagnetic wave and they can travel through the vacuum in outer space.
Explanation:
maturity also get disorganised
The definition of average acceleration allows to find the result for the average acceleration in the given time interval is:
Instantaneous acceleration is defined as the derivative of velocity with respect to time.
a = 
Where a is the acceleration, v the velocity and t the time.
They indicate that the speed of the car is given by the relation.
v = α t + β t²
With α = 3 m / s and β = 0.1 m / s³
Let's make the derivative.
a = α + 2β t
Let's substitute
a = 3 + 2 0.1 t
Average acceleration is the change in velocity in the time interval.
Let's find the velocity at the indicated time.
For t = 5 s
v₅ = 3 + 0.1 5²
v₅ = 5.5 m / s
For t = 10 s
v₁₀ = 3 + 0.1 10²
v₁₀ = 13 m / s
Let's calculate the average acceleration.
In conclusion using the definition of mean acceleration we can find the result for the mean acceleration in the given time interval is:
1.5 m / s²
Learn more here: brainly.com/question/20057878
