Answer:
A vector is any quantity with both magnitude and direction. Other examples of vectors include a velocity of 90 km/h east and a force of 500 newtons straight down.
Explanation:
After reading through all of the above, I don't find a question that needs to be answered.
But I just want to say:
Yep. Uh huh. Fer sher. You are true. Words of higher veracity are unlikely to be found. Every word of that scenario and its description is accurate, and cannot be debated or disputed in any wise.
What's more, I agree, and I thank you for the points.
Answer:
Explanation:
Resistance of wire, R = 48 ohm
It is cut into four pieces.
As we know that the resistance is directly proportional to the length of the wire.
So, the resistance of each piece of wire, R' = 48 / 4 = 12 ohm
Now the four pieces are connected in parallel. So the equivalent resistance is given by
![\frac{1}{R_{eq}}=\frac{1}{R'}+\frac{1}{R'}+\frac{1}{R'}+\frac{1}{R'}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Beq%7D%7D%3D%5Cfrac%7B1%7D%7BR%27%7D%2B%5Cfrac%7B1%7D%7BR%27%7D%2B%5Cfrac%7B1%7D%7BR%27%7D%2B%5Cfrac%7B1%7D%7BR%27%7D)
![\frac{1}{R_{eq}}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR_%7Beq%7D%7D%3D%5Cfrac%7B1%7D%7B12%7D%2B%5Cfrac%7B1%7D%7B12%7D%2B%5Cfrac%7B1%7D%7B12%7D%2B%5Cfrac%7B1%7D%7B12%7D)
Req = 3 ohm
Answer:
If the third charge is placed at a distance
the net force on the charge will be zero
Explanation:
<em>Consider the diagram below.</em>
<em>Let us add a new charge Q at a distance b from the origin.</em>
<em />
Let the force that the charge 2Q exert on the new charge (Q) be
![F1 = \frac{k2Q^{2} }{(a+b)^{2} }](https://tex.z-dn.net/?f=F1%20%3D%20%5Cfrac%7Bk2Q%5E%7B2%7D%20%7D%7B%28a%2Bb%29%5E%7B2%7D%20%7D)
Let the force that charge -Q exerts on the new charge be
![F2= \frac{-kQ^{2} }{b^{2} }](https://tex.z-dn.net/?f=F2%3D%20%5Cfrac%7B-kQ%5E%7B2%7D%20%7D%7Bb%5E%7B2%7D%20%7D)
For the third charge in equilibrium,
![\\F1 + F2 = 0\\\\\frac{k2Q^{2} }{(a+b)^{2} } - \frac{kQ^{2} }{(b)^{2} } =0](https://tex.z-dn.net/?f=%5C%5CF1%20%2B%20F2%20%3D%200%5C%5C%5C%5C%5Cfrac%7Bk2Q%5E%7B2%7D%20%7D%7B%28a%2Bb%29%5E%7B2%7D%20%7D%20-%20%5Cfrac%7BkQ%5E%7B2%7D%20%7D%7B%28b%29%5E%7B2%7D%20%7D%20%3D0)
Solving this will give us a quadratic equation which will be further simplified.
![\frac{k2Qx^{2}b ^{2}-kQ^{2}(a+b)^{2} }{(a+b)^{2}b^{2} }=0\\k2Qx^{2}b ^{2}-kQ^{2}(a+b)^{2} = 0\\\\kQx^{2}(2b ^{2}-(a+b)^{2}) = 0\\((2b ^{2}-(a+b)^{2}) = 0\\\\(2b ^{2}-a^{2}-2ab-b^{2}=0\\b^{2}-a^{2} -2ab=0\\a^{2} -b^{2} + 2ab =0\\\\\\](https://tex.z-dn.net/?f=%5Cfrac%7Bk2Qx%5E%7B2%7Db%20%5E%7B2%7D-kQ%5E%7B2%7D%28a%2Bb%29%5E%7B2%7D%20%20%20%7D%7B%28a%2Bb%29%5E%7B2%7Db%5E%7B2%7D%20%20%7D%3D0%5C%5Ck2Qx%5E%7B2%7Db%20%5E%7B2%7D-kQ%5E%7B2%7D%28a%2Bb%29%5E%7B2%7D%20%20%3D%200%5C%5C%5C%5CkQx%5E%7B2%7D%282b%20%5E%7B2%7D-%28a%2Bb%29%5E%7B2%7D%29%20%20%3D%200%5C%5C%28%282b%20%5E%7B2%7D-%28a%2Bb%29%5E%7B2%7D%29%20%20%3D%200%5C%5C%5C%5C%282b%20%5E%7B2%7D-a%5E%7B2%7D-2ab-b%5E%7B2%7D%3D0%5C%5Cb%5E%7B2%7D-a%5E%7B2%7D%20%20-2ab%3D0%5C%5Ca%5E%7B2%7D%20-b%5E%7B2%7D%20%2B%202ab%20%3D0%5C%5C%5C%5C%5C%5C)
Making b the subject of the formula, we can get the distance that the third charge will be placed .
![b=\sqrt{a^{2} +2ab}](https://tex.z-dn.net/?f=b%3D%5Csqrt%7Ba%5E%7B2%7D%20%2B2ab%7D)
So if the third charge is placed at a distance
the net force on the charge will be zero
<em>if real values were plugged in, it will be easier for us to determine the value of b</em>
<em />