All categories

2 years ago

What are some advantages of making electronic components like transistors increasingly smaller?

Engineering

1 answer:

Stells [14]2 years ago

4 0

Answer:

Dr. Engelbart, who would later help develop the computer mouse and other personal computing technologies, theorized that as electronic circuits were made smaller, their components would get faster, require less power and become cheaper to produce — all at an accelerating pace

You might be interested in

Type the correct answer in the box. Spell all words correctly. Mike is constructing a project in which he uses a motor. How can

tankabanditka [31]

If it is. DC, direct current reverse the polarity of power leads on the motor.

If it is a 3 phase ac alternating current, reverse any of the two of three leads.

Disconnect power before attempting.

6 0

3 years ago

Engineering Careers Scavenger Hunt

emmasim [6.3K]

Answer:

Explanation:

it's the only engineering career

6 0

3 years ago

Read 2 more answers

Does anybody know what plane this is? i saw it the other day doing a low pass through my community

Arlecino [84]

Answer:

Airbus A340-313

Explanation:

it is what it is

3 0

2 years ago

Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the

alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

$\dot Q_{ref} - \dot Q_{w} = 0$

$\dot Q_{ref} = \dot Q_{w}$

$\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})$

The mass flow rate of the cooling water is now cleared:

$\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}$

Given that $h_{ref,in} = 808.34\,\frac{kJ}{kg}$ , $h_{ref, out} = 88.82\,\frac{kJ}{kg}$ , $h_{w,out} = 104.83\,\frac{kJ}{kg}$ and $h_{w,in} = 62.98\,\frac{kJ}{kg}$ , the mass flow of the cooling water is:

$\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)$

$\dot m_{w} = 123.788\,\frac{kg}{min}$

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

4 0

3 years ago

List at least two things that scientists and explorers have in common.

Tresset [83]

• educated
•very curious
•discover new things
•Investigate
•search unknown

8 0

3 years ago

Read 2 more answers