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Anettt [7]
3 years ago
8

6. In an integrated circuit, each wafer is cut into sections, which

Physics
1 answer:
Katen [24]3 years ago
7 0

Answer:

B. carry a single circuit and are placed in individual cases.

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

Similarly, an integrated circuit (IC) also referred to as microchip can be defined as a semiconductor-based electronic component that comprises of many other tiny electronic components such as capacitors, resistors, transistors, and inductors.

Integrated circuits (ICs) are often used in virtually all modern electronic devices to carry out specific tasks or functions such as amplification, timer, oscillation, computer memory, microprocessor, etc.

A wafer can be defined as a thin slice of crystalline semiconductor such as silicon and germanium used typically for the construction of an integrated circuit.

In an integrated circuit, each wafer is cut into sections, which generally comprises of a single circuit that are placed in individual cases.

Additionally, a semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity.

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Can someone solve this problem and explain to me how you got it​
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1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

1)

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

r' = 5r

Therefore, the new force between the charges is:

F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

F=k\frac{q_1 q_2}{r^2}

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

- The charges are both changed by a factor of 8:

q_1' = 8q_1

q_2' = 8q_2

Substituting into the equation, we find the new force:

F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
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