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Sedaia [141]
3 years ago
10

5/6 - 1/3 sorry for add part i was meant to do minus but tell me what it is?​

Physics
2 answers:
bulgar [2K]3 years ago
5 0

Answer:

\frac{1}{2}

Explanation:

\frac{5}{6} -\frac{1}{3} = \frac{15}{18} -\frac{6}{18}=\frac{9}{18} =\frac{1}{2}

Gennadij [26K]3 years ago
4 0

Answer:

1/2

Explanation:

we will take the LCM of denominators with is LCM (6,3)=6.

Now we shall make denominators equal no both sides. Since 5/6 already has 6 as denominator we shall multiply 2 with 3 to make denominators equal in 1/3 and so we also multiply 1 with 2. we get

1/3 × 1/2= 1/6 ×2/1=2/6

now 5/6-2/6=3/6=1/2

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I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.

<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know V and I use formula P = IV: P = IV = (100mA)(10V) = 1 W.</span>

The next question is what I'm not sure about:

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What I did initially was: P = IV = (100mA)(2V) = 2 W

But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."

So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.

P = IV = (200mA)(2V) = 4 W

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