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vova2212 [387]
2 years ago
6

PLEASE HELP QUICK!!

Engineering
1 answer:
ivolga24 [154]2 years ago
5 0

R01= 14.1 Ω

R02=  0.03525Ω

<h3>Calculations and Parameters</h3>

Given:

K= E2/E1 = 120/2400

= 0.5

R1= 0.1 Ω, X1= 0.22Ω

R2= 0.035Ω, X2= 0.012Ω

The equivalence resistance as referred to both primary and secondary,

R01= R1 + R2

= R1 + R2/K2

= 0.1 + (0.035/9(0.05)^2)

= 14.1 Ω

R02= R2 + R1

=R2 + K^2.R1

= 0.035 + (0.05)^2 * 0.1

= 0.03525Ω

Read more about resistance here:

brainly.com/question/17563681

#SPJ1

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A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir
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Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

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So, internal radius, r = 230 mm

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Density of molten metal, ρ = $7.2 \ g/cm^3$

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The height of pouring cavity above parting surface is h = 300 mm

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So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

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3 0
3 years ago
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