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.
Answer:
= 8.33 Watt
Explanation:
where,
p = resistivity
l = length
A = cross section area
Given that ,
p = resistivity = 6.0 × 10–8 Ω
l = 2m
A = cross section area = 2.0 mm × 2.0 mm = 4 x 10^-6 m^2
A = 2 x 2 mm^2 = 4 x 10^-6 m^2
p = 6 x 10^-8 ohm metre,
V = 0.5 V
Let R be the resistance of the rod
R = 3 × 10⁻²Ω
Heat generated = V^2 / R
= (0.5)^2 / (3 x 10^-2)
= 8.33 Watt
The correct matches are as follows:
<span>1. Mechanical Energy
</span>This is the energy of motion and position - Kinetic Energy and Potential Energy are included here.
<span>
2. Electrical Energy
</span>This is the type of energy that is given off as a result of electrons (charges!) moving through a conductor.
<span>
3. Light Energy
</span>This is the only form of energy we can see with our eyes!
<span>
4. Thermal Energy
</span>This is heat energy! It can be transferred through conduction, convection, or radiation.
<span>
5. Sound Energy
</span>This is energy that is transmitted through vibrations.
<span>
6. Kinetic Energy
</span>This is the energy that all moving objects have. <span>
7. Potential Energy
</span>This is stored energy, waiting to be used!
The impact force makes sense because the impulse experienced by the body cause a great change in momentum of the body.
<h3 /><h3>What is impulse?</h3>
This is the force that acts on a body over a given period of time. The impulse experienced by a body is determined as the product of force and time of action.
J = Ft
The change in momentum of a body is equal to the impulse experienced by the body.
ΔP = Ft
Thus, the impact force makes sense because the impulse experienced by the body cause a great change in momentum of the body.
Learn more about impulse here: brainly.com/question/25700778
Answer:
The force bumper at 0.200m
F=2722.5 N
Explanation:
Using the energy theorem of work
W=ΔK
W=F*d
ΔK=
ΔK=F*d=
