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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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Consider a potato being baked in an oven that is maintained at a constant temperature. the temperature of the potato is observed
solong [7]

Answer:

<em>The rise in temperature will be less than 5 °C in the second minute.</em>

Explanation:

According to heat conduction law, the rate of heating is proportional to the temperature difference or temperature gradient. The temperature gradient is what drives heat to move from a hotter body at a higher temperature gradient to a colder body at a lower temperature gradient. For the potato, the initial first minute raises the temperature to 5 °C, consequently reducing the temperature gradient between the potato and the heating element in the oven. <em>This reduced temperature gradient means that the rate at which it will conduct heat in the second minute will be lesser than that at the first minute</em>. This will continue till the potato and the heating element are at the same temperature, at which no temperature gradient will exist between them; stopping heat transfer between them.

3 0
3 years ago
A master stud pattern is laid out somewhat<br> like a?
Svetradugi [14.3K]

Answer:

••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.

3 0
2 years ago
Calculate the equivalent capacitance of the three series capacitors in Figure 12-1
GrogVix [38]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

a) 0.01 μF

b) 0.58 μF

c) 0.060 μF

d) 0.8 μF

Answer:

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

Explanation:

Please refer to the attached Figure 12-1 where three capacitors are connected in series.

We are asked to find out the equivalent capacitance of this circuit.

Recall that the equivalent capacitance in series is given by

$ \frac{1}{C_{eq}} =  \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} $

Where C₁, C₂, and C₃ are the individual capacitance connected in series.

C₁ = 0.1 μF

C₂ = 0.22 μF

C₃ = 0.47 μF

So the equivalent capacitance is

$ \frac{1}{C_{eq}} =  \frac{1}{0.1} + \frac{1}{0.22} + \frac{1}{0.47} $

$ \frac{1}{C_{eq}} =  \frac{8620}{517}  $

$ C_{eq} =  \frac{517}{8620}  $

$ C_{eq} =  0.0599  $

Rounding off yields

$ C_{eq} =  0.060 \: \mu F $

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

5 0
3 years ago
A crude fermenter is set up in a shed in the backyard of a suburban house. Under anaerobic conditions with ammonia as the nitrog
Aleks04 [339]

Answer:

using calculations Heat losses will be 4512 J

5 0
3 years ago
A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N
soldier1979 [14.2K]

Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, P_{t} = 8 kW = 8000 W

Now, the force, F is given by:

P_{t} = Force(F)\times velocity(v) = Fv               (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, F_{c} is provided by the gravitational force, F_{G}:

F_{c} = F_{G}

\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}

Thus from the above, velocity comes out to be:

v = \sqrt{\frac{GM_{e}}{R}}

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s

where

R = R_{e} + H

R = \sqrt{GM_{e}(\frac{T}{2\pi})^{2}}

where

G = Gravitational constant

T = Time period of rotation of Earth

R is calculated as 42166 km

Now, from eqn (1):

8000 = F\times 3075.36

F = 2.60 N

6 0
3 years ago
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