need some help answering these questions

Question

anonymous · Answer

attachment

anonymous · Answer

im not to good with this stuff

anonymous · Answer

wat s the question actually
?????????????????

anonymous · Answer

its in the pdf

anonymous · Answer

it has a picture so i couldnt post it

anonymous · Answer

1) $$9 \times 10^{9}$$

anonymous · Answer

$$2)  (1.6 \times10^{-19})^{2} N$$

anonymous · Answer

3)all the resistors are in parallel so ,$$r _{p}=2\Omega$$

anonymous · Answer

v=ir
=>9 = i(2)
=>i=9/2= 4.5 A

anonymous · Answer

p = i^2 r
=>p=(9/2)^2 (2)
=>p= 81/2
=>p=40.5

anonymous · Answer

@zerocool127 : is these ur answers

anonymous · Answer

Firstly, this would be better placed in the Physics area.

Use Coulomb's law $$F = kq _{1}q _{2}/r ^{2}$$

The resistors are in parallel, so 1 over the total is equal to the sums of the inverses.
$$1/R = 1/R _{1} + 1/R _{2} + 1/R _{3} $$
R = 1/(1/5 + 1/5 + 1/10) = 2 ohms

V = IR   I = V/R  =  9/2 A  =  4.5A

P = j/s = j/q * q/s = VA = 4.5*9W = 40.5W

anonymous · Answer

that was my ans toooooooooooo

anonymous · Answer

DHASHNI, I saw you were doing the first one, so I tried to finish up the others, but ended up taking too long. Engineering major?

anonymous · Answer

no!!!

anonymous · Answer

is there any other question?????