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maks197457 [2]
3 years ago
11

A certain computer chip that has dimensions of 3.67 cm and 2.93 cm contains 3.5 million transistors. If the transistors are squa

re, what must be their maximum dimensions (in terms of microns or µm)? (Note: Devices other than transistors are also on the chip, and there must be room for the interconnections among the circuit elements. Transistors smaller than 0.7 μm are now commonly and inexpensively fabricated.)
Physics
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer:

1.56 × 10^-3 cm.

Explanation:

So, we are given the following parameters from the question above;

Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.

Step one: find the area of the computer chip.

Therefore, Area = Length × breadth.

Area = 3.67 cm × 2.93 cm.

Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.

Step two: find the area of one transistor

The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).

Hence;

The area of one transistor= 10.7531/4.4 × 10^6.

The area of one transistor= 2.44 × 10^-6 cm^2.

=> Note that We have our transistors as square, therefore;

The maximum dimension = √ (2.44 × 10^-6) cm^2.

The maximum dimension= 1.56 × 10^-3 cm.

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Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black
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\sigma = Stefan-Boltzmann constant

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The view factor between two coaxial parallel disks would be

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Then the view factor between base to top surface of the cylinder becomes F_{12} = 0.26. From the summation rule

F_{13} = 1-0.26

F_{13} = 0.74

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}

\dot{Q_3} = 2\dot{Q_{13}}

\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)

\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)

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Read 2 more answers
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The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

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The average speed of the ant is calculated as;

average \ speed = \frac{total \ distance }{total \ time }

The total distance from the graph is calculated as follows;

  • first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm
  • first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm
  • second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm
  • second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm
  • third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm
  • fourth downward distance from 12 cm to 9 cm = 3 cm
  • final horizontal distance from 13 cm to 15 cm = 2cm

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

  • This shortest distance is the straight line connecting the start and end position. Call this line P
  • From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm
  • Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of  line P.

                                                ↓end

                                                ↓

                                                ↓ 6cm

                                                ↓

  start -------------13 cm------------

Use Pythagoras theorem to solve for P.

P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm

The average velocity of the ant is calculated as;

average \ velocity= \frac{\Delta displacemnt  }{total\ time }= \frac{14.318 \ cm}{105 \  s} = 0.136 \ cm/s  \\\\

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

Learn more here: brainly.com/question/589950

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2 years ago
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