Answer:
The value of charge q₃ is 40.46 μC.
Explanation:
Given that.
Magnitude of net force ![F=14.413\ N](https://tex.z-dn.net/?f=F%3D14.413%5C%20N)
Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.
We need to calculate the distance
Using Pythagorean theorem
![r=\sqrt{x_{2}^2+y_{2}^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7Bx_%7B2%7D%5E2%2By_%7B2%7D%5E2%7D)
Put the value into the formula
![r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%288.2%5Ctimes10%5E%7B-2%7D%29%5E2%2B%283.1%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![r=0.0876\ m](https://tex.z-dn.net/?f=r%3D0.0876%5C%20m)
We need to calculate the magnitude of the charge q₃
Using formula of net force
![F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})](https://tex.z-dn.net/?f=F_%7B12%7D%3Dkq_%7B2%7D%28%5Cdfrac%7Bq_%7B3%7D%7D%7Br_%7B3%7D%5E2%7D%2B%5Cdfrac%7Bq_%7B1%7D%7D%7Br_%7B1%7D%5E2%7D%29)
Put the value into the formula
![14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})](https://tex.z-dn.net/?f=14.413%3D9%5Ctimes10%5E%7B9%7D%5Ctimes7.7%5Ctimes10%5E%7B-6%7D%28%5Cdfrac%7Bq_%7B3%7D%7D%7B%280.0438%29%5E2%7D%2B%5Cdfrac%7B-3%5Ctimes10%5E%7B-6%7D%7D%7B%280.0876%29%5E2%7D%29)
![(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}](https://tex.z-dn.net/?f=%28%5Cdfrac%7Bq_%7B3%7D%7D%7B%284.38%5Ctimes10%5E%7B-2%7D%29%5E2%7D%2B%5Cdfrac%7B-3%5Ctimes10%5E%7B-6%7D%7D%7B%280.0876%29%5E2%7D%29%3D%5Cdfrac%7B14.413%7D%7B9%5Ctimes10%5E%7B9%7D%5Ctimes7.7%5Ctimes10%5E%7B-6%7D%7D)
![\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}](https://tex.z-dn.net/?f=%5Cdfrac%7Bq_%7B3%7D%7D%7B%280.0438%29%5E2%7D%3D207%5Ctimes10%5E%7B-4%7D%2B3.909%5Ctimes10%5E%7B-4%7D)
![q_{3}=0.0210909\times(0.0438)^2](https://tex.z-dn.net/?f=q_%7B3%7D%3D0.0210909%5Ctimes%280.0438%29%5E2)
![q_{3}=40.46\times10^{-6}\ C](https://tex.z-dn.net/?f=q_%7B3%7D%3D40.46%5Ctimes10%5E%7B-6%7D%5C%20C)
![q_{3}=40.46\ \mu C](https://tex.z-dn.net/?f=q_%7B3%7D%3D40.46%5C%20%5Cmu%20C)
Hence, The value of charge q₃ is 40.46 μC.
The convection going on with the magma in the asthenosphere
Answer:
The factor of the diameter is 0.95.
Explanation:
Given that,
Power of old light bulb = 54.3 W
Power = 60 W
We know that,
The resistance is inversely proportional to the diameter.
![R\propto\dfrac{1}{D}](https://tex.z-dn.net/?f=R%5Cpropto%5Cdfrac%7B1%7D%7BD%7D)
The power is inversely proportional to the resistance.
![P\propto\dfrac{1}{R}](https://tex.z-dn.net/?f=P%5Cpropto%5Cdfrac%7B1%7D%7BR%7D)
![P\propto D^2](https://tex.z-dn.net/?f=P%5Cpropto%20D%5E2)
We need to calculate the factor of the diameter of the filament reduced
Using relation of power and diameter
![\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}](https://tex.z-dn.net/?f=%5Cdfrac%7BP_%7Bi%7D%7D%7BP_%7Bf%7D%7D%3D%5Cdfrac%7BD_%7Bi%7D%5E2%7D%7BD_%7Bf%7D%5E2%7D)
Put the value into the formula
![\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}](https://tex.z-dn.net/?f=%5Cdfrac%7BD_%7Bi%7D%5E2%7D%7BD_%7Bf%7D%5E2%7D%3D%5Cdfrac%7B54.3%7D%7B60%7D)
![\dfrac{D_{i}}{D_{f}}=0.95](https://tex.z-dn.net/?f=%5Cdfrac%7BD_%7Bi%7D%7D%7BD_%7Bf%7D%7D%3D0.95)
![D_{i}=0.95 D_{f}](https://tex.z-dn.net/?f=D_%7Bi%7D%3D0.95%20D_%7Bf%7D)
Hence, The factor of the diameter is 0.95.
The magnetic field is described mathematically as a vector field<span>. This vector field can be plotted directly as a set of many vectors drawn on a grid. Each vector points in the direction that a compass would point and has length dependent on the strength of the magnetic force. </span>
Answer:
The current will decrease.
Explanation:
When another bulb is added, the resistance is going to increase. Keep in mind that the current is inversely proportional to the resistance (<em>Ohm's law: R= </em><em>V</em><em>/</em><em>I</em><em> </em><em>).</em> Therefore when the resistance increase, the current running in the circuit will decrease.