A camera flash or lighting bolt because stored separated positive and negative charges --> caused them to do work by briefly lighting a bulb as the separated charges moved back together
Answer:In a DC circuit, the power consumed is simply the product of the DC voltage times the DC current, given in watts.for AC circuits with reactive components we have to calculate the consumed power differently.
a 1/4 watt resistor or a 20 watt amplifier.
First electromagnet
Explanation:
The first electromagnet is the strongest and it is stronger than the given electromagnet above.
An electromagnet is a temporary magnet made by passing current through a wire wounded round an iron core or metallic core.
- When current is passed through, the metal becomes magnetic.
- The strength of the electromagnet depends on the number of coil round the metal core and also the intensity of current passed through it.
- The higher the number of coils wounded round the metal core, the stronger the electromagnet that will be produced.
- Also, the higher the intensity of electricity passed through the wire, the stronger it is.
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Electromagnet brainly.com/question/2191993
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This question is incomplete, the complete question is;
Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).
how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of charge and the distance between the charges. There are many correct answers
Answer:
Given the data in question;
Dipole moment P = 1 × 10⁻⁹ C.m
now dipole pointing to the right;
P→
(-) ---------------->(+) 
d
so let distance between the dipoles be d
∴ P = d
Let
= 1 nC
so
P = d
1 × 10⁻⁹ = 1 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (1 × 10⁻⁹)
d = 1 m
Also Let
= 2 nC
so
P = d
1 × 10⁻⁹ = 2 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (2 × 10⁻⁹)
d = 0.5 m
Also Let
= 3 nC
so
P = d
1 × 10⁻⁹ = 3 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (3 × 10⁻⁹)
d = 0.33 m
such that;
charge distance
1 nC 1.00 m
2 nC 0.50 m
3 nc 0.33 m
4 nC 0.25 m
5 nC 0.20 m