Everything is in the universe.
That's what the word means.
The equation for electrical power is<span>P=VI</span>where V is the voltage and I is the current. This can be rearranged to solve for I in 6(a).
6(b) can be solved with Ohm's Law<span>V=IR</span>or if you'd like, from power, after substituting Ohm's law in for I<span>P=<span><span>V2</span>R</span></span>
For 7, realize that because they are in parallel, their voltages are the same.
We can find the resistance of each lamp from<span>P=<span><span>V2</span>R</span></span>Then the equivalent resistance as<span><span>1<span>R∗</span></span>=<span>1<span>R1</span></span>+<span>1<span>R2</span></span></span>Then the total power as<span><span>Pt</span>=<span><span>V2</span><span>R∗</span></span></span>However, this will reveal that (with a bit of algebra)<span><span>Pt</span>=<span>P1</span>+<span>P2</span></span>
For 8, again the resistance can be found as<span>P=<span><span>V2</span>R</span></span>The energy usage is simply<span><span>E=P⋅t</span></span>
Velocity is defined as a change in position.
Answer:
The magnitude of electric force is 
Explanation:
Coulomb's Law:
The force of attraction or repletion is
- directly proportional to the products of charges i.e

- inversely proportional to the square of distance i.e


[ k is proportional constant=9×10⁹N m²/C²]
There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C
Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C
and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C
Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.
If we draw a line from q₁ to Q .
The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.
Let hypotenuse = r
Therefore, 
we know,


Total force 


[ r=5]
N
The magnitude of electric force is 