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Rufina [12.5K]
3 years ago
12

2.1 What is the minimum number of pins required for a so-called dual-op-amp IC package, one containing two op amps? What is the

number of pins required for a so-called quad-op-amp package, one containing four op amps?
Engineering
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

8 for dual-op-amp package, and 14 for quad-op-amp

Explanation;

This is because every op-amp has 2 input terminal 4 pns

So one output terminal that is 2 pins which are required for power

and the same for a minumum number of pins required by quad op amp which is 14

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Hello, I have a question, I would be glad if you can help.
Anastasy [175]

Answer:

  infinite

Explanation:

The tangent of the angle the arm makes with the vertical will be the ratio of the centripetal acceleration to the acceleration due to gravity on the center of mass of the arm. The angle can only be 90° (fully-open arms) if that ratio is infinite.

The speed must be infinite for the arms to be fully open.

3 0
3 years ago
A 0.50 m3 drum was filled with 0.49 m3 of liquid water at 25oC and the remaining volume was water vapor without any air. The dru
timurjin [86]

Answer:

There is not going to be pressure build up in the  system,that is isobaric process.

Explanation:

Assumptions to be made

1.  No mass is gained or lost during the heating process.

2. There are no friction losses,so work is transmitted efficiently.

3. It was started the water in the drum and its surrounding have same temperature.

4. This system is closed,so there is no mass transfer across its boundaries.

5 0
3 years ago
Re armature of a 4 pole DC generator is required to generate an emf of 520v on open circuit when revolving at a speed of 660rpm.
bija089 [108]

Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.

<u>Given the following data:</u>

  • Emf = 520 Volts
  • Speed = 660 r.p.m
  • Number of armature conductors = 144 slots
  • Number of poles = 4 poles
  • Number of parallel paths = 2

To find the magnetic flux per pole:

Mathematically, the emf generated by a DC generator is given by the formula;

E = \frac{\theta ZN}{60} × \frac{P}{A}

<u>Where:</u>

  • E is the electromotive force in the DC generator.
  • Z is the total number of armature conductors.
  • N is the speed or armature rotation in r.p.m.
  • P is the number of poles.
  • A is the number of parallel paths in armature.
  • Ф is the magnetic flux.

First of all, we would determine the total number of armature conductors:

Z = 144 × 2 × 3

Z = 864

Substituting the given parameters into the formula, we have;

520 = \frac{\theta (864)(660)}{60} × \frac{4}{2}

520 = \theta (864)(11) × 2

520 = 19008 \theta \\\\\Theta = \frac{520}{19008}

<em>Magnetic flux </em><em>=</em><em> 0.0274 Weber.</em>

Therefore, the magnetic flux per pole is 0.0274 Weber.

Read more: brainly.com/question/15449812?referrer=searchResults

5 0
3 years ago
a metal rod 24mm diameter and 2m long is subjected to an axial pull of 40 kN. If the rod is 0.5mm, then find the stress-induced
ozzi

Answer:

i dont know but i will take the points tho hahah

Explanation:

8 0
3 years ago
Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire
Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0
3 years ago
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