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>
Answer:
10250 N/C leftwards
Explanation:
QA = 4 micro Coulomb
QB = - 5 micro Coulomb
AP = 6 m
BP = 2 m
A is origin, B is at 4 m and P is at 6 m .
The electric field due to charge QA at P is EA rightwards
The electric field due to charge QB at P is EB leftwards
The resultant electric field at P due the charges is given by
E = EB - EA
E = 11250 - 1000 = 10250 N/C leftwards
The diagram is showing a 3d model of an atom, with all of the electrons demonstrated in a rounded shape, which resembles a cloud, thus being called an electron cloud.
There are usually multiple methods for organizing items and information in a scientific investigation.
The primary colors of light are red, blue and green.
There are the pigments like yellow, magenta and cyan that are the mixture of two primary colors.
For example, magenta is a mixture of red and blue color. Thus, it reflects the red and blue color. Also, magneta absorbs the green color.
These type of colors that reflects two primary colors and absorb one color are known as secondary pigments.
Hence, 2nd option is the correct answer.
