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Ilia_Sergeevich [38]
3 years ago
7

What are the best collages for architectural learning?

Engineering
1 answer:
Sati [7]3 years ago
4 0

Answer:

I don't where you live, but I named few around the world. Some are Virgi..nia Tech, Unive..rsity of Flor..dia, and Oklah..oma Stat..e Univ..ersity.

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A 55-μF capacitor has energy ω (t) = 10 cos2 377t J and consider a positive v(t). Determine the current through the capacitor.
mart [117]

Given :

Capacitor , C = 55 μF .

Energy is given by :

\omega(t)=10cos^2 (377t)\ J .

To Find :

The current through the capacitor.

Solution :

Energy in capacitor is given by :

\omega=\dfrac{Cv^2}{2}\\\\v=\sqrt{\dfrac{2\omega}{C}}\\\\v=\sqrt{\dfrac{2\times 10cos^2 (377t)}{55\times 10^{-6}}}\\\\v=cos(337t)\sqrt{\dfrac{2\times 10}{55\times 10^{-6}}}\\\\v=603.02\ cos( 337t)

Now , current i is given by :

i=C\dfrac{dv}{dt}\\\\i=C\dfrac{d[603.02cos(337t)]}{dt}\\\\i=-55\times 10^{-6}\times 603.03\times 337\times sin(337t)\\\\i=-11.18\ sin(337t)

( differentiation of cos x is - sin x )

Therefore , the current through the capacitor is -11.18 sin ( 377t).

Hence , this is the required solution .

6 0
3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa . It has been de
goblinko [34]

Answer:

a fracture will occur, because the Kc value is greater than the KIC (48.9901 MPa > 40 MPa)

Explanation:

the solution is in the attached Word file

Download docx
7 0
3 years ago
On some engines, after torquing cylinder head fasteners, you
djverab [1.8K]

Answer:

Re-torque the bolts as required while your engine is warm. But if you're using aluminum cylinder heads, you should wait until your engine is complete cooled until re-torquing

3 0
3 years ago
A piston-cylinder device contains 1.329 kg of nitrogen gas at 120 kPa and 27 degree C. The gas is now compressed slowly in a pol
Shtirlitz [24]

Answer:-0.4199 J/k

Explanation:

Given data

mass of nitrogen(m)=1.329 Kg

Initial pressure\left ( P_1\right )=120KPa

Initial temperature\left ( T_1\right )=27\degree \approx 300k

Final volume is half of initial

R=particular gas constant

Therefore initial volume of gas is given by

PV=mRT

V=0.986\times 10^{-3}

Using PV^{1.49}=constant

P_{1}V^{1.49}=P_2\left (\frac{V}{2}\right )

P_2=337.066KPa

V_2=0.493\times 10^{-3} m^{3}

and entropy is given by

\Delta s=C_v \ln \left (\frac{P_2}{P_1}\right )+C_p \ln \left (\frac{V_2}{V_1}\right )

Where, C_v=\frac{R}{\gamma-1}=0.6059

C_p=\frac{\gamma R}{\gamma -1}=0.9027

Substituting values we get

\Delta s=0.6059\times\ln \left (\frac{337.066}{120}\right )+0.9027 \ln \left (\frac{1}{2}\right )

\Delta s=-0.4199 J/k

4 0
4 years ago
Which of the following is a way to heat or cool a building without using electricity or another power source?
zzz [600]
I believe the answer is: A. Passive heating and cooling.
8 0
3 years ago
Read 2 more answers
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