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Dvinal [7]
3 years ago
7

Answer every question of this quiz

Engineering
1 answer:
Reil [10]3 years ago
5 0

I'd say number 4, number 3 looks like an exhaust valve

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Suppose the loop is moving toward the solenoid (to the right). Will current flow through the loop down the front, up the front,
Tems11 [23]

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See explanation

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6 0
2 years ago
Read 2 more answers
Basic concepts surrounding electrical circuitry?​
Elanso [62]
Hopefully that helps you out and is this for history or science?

3 0
2 years ago
For a column that is pinned at both ends, the critical buckling load can be calculated as, Pcr = π2 E I /L^2 where E is Young's
gulaghasi [49]

When a slender member is subjected to an axial compressive load, it may fail by a ... Consider a column of length, L, cross-sectional Moment of Inertia, I, having Young's Modulus, E. Both ends are pinned, meaning they can freely rotate ... p2EI L2 ... scr, is the Euler Buckling Load divided by the columns cross-sectional area

6 0
3 years ago
What type of siege engines were used by Saladin to capture Jerusalem in 1187?
mariarad [96]

Answer:

LOTS

Explanation:

Catapults, Towers, and Trebuchets were all used by Saladin to capture Jerusalem in 1187

8 0
2 years ago
Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier.
atroni [7]

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage v_p=120volt

(i) We know that peak voltage is give by v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt

(ii) We know that for transformer \frac{v_p}{v_s}=\frac{n_p}{n_s}

So \frac{169.08}{v_s}=\frac{10}{1}

v_s=16.90volt

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt

(v) Now dc current is given by i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A

4 0
2 years ago
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