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Bas_tet [7]
1 year ago
12

- Light bulbs can be used to indicate current flow in a circuit. The brightness of a bulb is proportional to the amount of curre

nt passing throw it. The figure shows a battery, a switch, two light bulbs, and a capacitor that is initially uncharged
Physics
1 answer:
Rama09 [41]1 year ago
5 0

Light bulbs can be used to indicate current flow in a circuit. The brightness of a bulb is proportional to the amount of current passing throw it. The figure shows a battery, a switch, two light bulbs, and a capacitor that is initially uncharged The brightness of a lightbulb is given by its power. P = I2R, and so brightness depends on current and resistance

<h3>What is current flow?</h3>

A stream of charged particles, such as electrons or ions, traveling through an electrical conductor or a vacuum is known as an electric current. The net rate of electric charge flowing through a surface or into a control volume is how it is calculated. [622 Charge carriers, which can be any of a number of particle kinds depending on the conductor, are the moving particles. Electrons flowing over a wire are frequently used as charge carriers in electric circuits. They can be electrons or holes in semiconductors. Ions are the charge carriers in an electrolyte, whereas ions and electrons are the charge carriers in plasma, an ionized gas.

To learn more about current flow from the given link:

brainly.com/question/22650781

#SPJ4

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A wave has a wavelength of 20 mm and a frequency of 5 Hz what is the speed?
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A box has a 20 N force applied to it to move it 5 m. What is the work done on the box? 4 J 4 N 25 J 100 J
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3 years ago
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Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0
3 years ago
Can anyone plz help me ​
juin [17]

Answer:

D,B,C,A,C

Explanation:

I believe that is the correct answers but it is unclear. I don't think the key for the second last question would let the current flowing so the bulb would be off.

8 0
3 years ago
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