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3 years ago

Which aesthetic principle do architects generally follow?

Engineering

1 answer:

Serjik [45]3 years ago

8 0

Answer:

C. Form follows function.

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An irreversible heat pump and a Carnot heat pump operate between the same two thermal reservoirs. Which heat pump has higher COP

Shkiper50 [21]

Answer:

Carnot heat pump

Explanation:

Carnot heat pump is an ideal heat pump in which all processes are reversible and that consume minimum amount of work to and produces maximum amount of heating effect compare to all real engine.And that is why COP of Carnot heat pump is more as compare to real heat pump.

All real heat pump are not perfectly reversible heat pump So this is also called irreversible heat pump .Due to irreversibility the COP of irreversible heat pump is always less than the COP of Carnot heat pump.

6 0

3 years ago

CAD software is indispensable for only creating engineering tools.<br> true or false

astraxan [27]

Answer:

False

Explanation:

CAD software is used for designing and making 3D models of almost anything you can imagine.

5 0

3 years ago

We know that passengers can be either helpful or harmful to a driver. Describe a pro and a con of having passengers in your car.

anygoal [31]

Answer:

sub to pewdipie

6 0

3 years ago

Read 2 more answers

Instrument panel help please

nirvana33 [79]

Answer:

A. 4WD Low

B. 4WD High

C. Safety belt

D. Check gauge

E. Oil temperature gauge

F. Left turn indicator

G. Tachometer

H. High beams

I. Speedometer

J. Right turn indicator

K. Battery voltage gauge

L. Cruise control light

M. (can't see sorry)

N. Service engine soon

O. Battery warning light

P. Airbag indicator

Q. Anti-lock brake system

R. Fuel gauge

S. Parking brake (AKA emergency brake) on

T. Low windshield washer fluid

U. Check fuel cap

V. Door ajar/open

W. Shift indicator

X. Trip odometer reset

Y. Oil pressure gauge

Z. compass/cardinal direction indicator

AA. Odometer?

4 0

3 years ago

At an operating frequency of 5 GHz, a 50 lossless coaxial line with insulating material having a relative permittivity r = 2.25

lapo4ka [179]

Answer:

The answer is "150 $\Omega$ ".

Explanation:

Its line length must be converted into wavelengths for the Smith chart to be used.

$\to \beta = \frac{2 \pi }{\lambda } \\\\\to u_p = \frac{\omega }{\beta}\\\\\lambda =\frac{2 pi}{\beta} =\frac{2 \pi U_p}{\omega}=\frac{c}{\sqrt{\varepsilon_r f}}$

$= \frac{3 \times 10^8}{ 2.25 \times (5 \times 10^9)}\\\\= \frac{3}{ 2.25 \times 5\times 10 }\\\\= \frac{3}{ 22.5 \times 5 }\\\\= \frac{3}{ 112.5}\\\\= 0.04 \ m.$

$l =\frac{0.30 }{0.04 } \times \lambda$

$= 7.5 \lambda$

Because it is an integrated half-wavelength amount,

$Z_{in} = Z_L = 150 \ \Omega$

8 0

3 years ago