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GREYUIT [131]
3 years ago
14

Which part of a circuit creates an electric force field that makes it possible for the circuit to work?

Physics
1 answer:
8090 [49]3 years ago
7 0
Every working circuit always includes a battery or other 'power supply'.

That component not only creates an electrostatic field within the conductors
in the circuit, but it also supplies energy to the circuit.
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Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st
Black_prince [1.1K]

Answer:

Part A:

E_{midpoint}=0

Part B:

E_{center}=2711.7558 N/C

Explanation:

Part A:

Formula of Electric Field Strength:

E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}

Where:

x is the distance from the ring

R is the radius of the ring

\epsilon is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C

Electric Field Strength at center of left ring is same as that of right ring.

E_{center}=2711.7558 N/C

5 0
3 years ago
The first step in turning a rock into a sediment is ________. Select one:
Debora [2.8K]

Answer:

The first step in turning a rock into a sediment is Compaction .

Explanation:

Lithification is a process of changing rock into sediments. There are two steps for a rock to lithify. These steps are as follows

  • The first step of lithification is compaction where sediments are erosed together by weight of the it. Thus, the upper layers of sediments causes compaction of lower layers.
  • The next process of lithification is cementation. In this fluids fill the space between the loose particles.

Hence, the first step for turning rock into sediments is compaction.


6 0
3 years ago
Determine the magnitude and direction of the resultant force of the following free body diagram.
Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force (\vec F), in newtons, by vectorial sum:

\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j} (1)

\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}

\|\vec F\| \approx 599.923\,N

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)

Where \theta is the direction of the resultant force, in sexagesimal degrees.

\theta \approx 36.405^{\circ}

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

4 0
3 years ago
Nuclear chain reactions within power plants do NOT produce bomb-like explosions primarily because the ________.
Gelneren [198K]

Answer: D. Density of uranium within nuclear fuel rods is insufficient to become explosive

Explanation: Nuclear power plants use the same fuel as nuclear bombs, i.e. radioactive Uranium-235 isotope. However, in a nuclear power plant, the energy is released more slowly unlike in a nuclear bomb. <em>The energy released is through nuclear fission, and radioactive decay occurs at the same rate as in nuclear bombs. therefore, option A, B</em><em> </em><em>and C are incorrect.</em>

The primary reason why nuclear chain reactions within power plants do NOT produce bomb-like explosions is because the uranium fuel rods used in electricity generation is not sufficiently enriched in Uranium-235 to produce a nuclear detonation. This is the same idea in option D which is the correct option.

6 0
3 years ago
The electrons involved in the formation of a chemical bond are called
tatiyna

Answer: Valence electrons

Valence electrons are those that are in the outermost or superficial layer of the atom, which means they have the highest energy compared to those of the inner layers.

Because of their position, it is easier for these electrons to interact with other atoms of their own element as well as different elements. This is done through the process of forming bonds when being attracted by other atoms.

7 0
3 years ago
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