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Aleksandr-060686 [28]
3 years ago
10

Ok, so I've answered this question but I really need someone to just check through it because I have this feeling that I've done

it wron

Physics
2 answers:
bonufazy [111]3 years ago
5 0
Your fill-ins for the first three lines of the table are correct.  Good work.

You have all the same answers for the last 2 lines of the table, like
it makes no difference to anything whether S2 is open or closed. 
That's not correct. It does make a difference.  If S2 is closed, then
the current would much rather go through S2 than through the bulbs
C and D because it's so much easier, and that's exactly what it does.
When S2 is closed, bulbs C and D are off. 

But I'll say again:  Except for those two middle boxes on the bottom line
of the table, everything else is correct.  That's 12 correct fill-ins for you,
and only 2 incorrect ones.  That's a score of 86% !  After posting so many
of these on Brainly this afternoon, doesn't it feel great to do your own work
for a change, and get it almost all correct ?
olga_2 [115]3 years ago
4 0
I believe you are correct
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When Magenta light is shown on a green surface, it looks black.

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3 years ago
If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w
marshall27 [118]

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

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