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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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Gennadij [26K]

Answer:

II

Explanation:

3 0
3 years ago
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical
Olin [163]

Answer:

The theoretical density for Niobium is 1.87 g/cm^3.

Explanation:

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density  of the unit cell

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm

1 nm = 10^{-7} cm

On substituting all the given values , we will get the value of 'a'.

\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}

\rho =1.87 g/cm^3

The theoretical density for Niobium is 1.87 g/cm^3.

6 0
3 years ago
Read 2 more answers
If the stopping sight distance required to avoid colliding with a fallen boulder on a mountain pass is 800 feet, what is the lik
Anettt [7]

Answer:

Grade is 10.32%

Explanation:

Given data:

stopping side distance is 800ft

t = 2.5 sec

design speed v = 65 miles/hr

a/g value is 0.35

stopping side distance is SSD

SSD = 1.47 vt + \frac{v^2}{30(a/g - G}

plugging all value for G

800 = 1.47 \times 65\times 2.4 + \frac{65^2}{30(0.35 -G)}

800 - 229.32 = \frac{65^2}{30(0.35 -G)}

570.68 = \frac{65^2}{30(0.35 -G)}

G = 0.1032

Grade is 10.32%

5 0
3 years ago
A 15,000 kg satellite is orbiting at 800 km above the Earth’s surface. What is the potential energy of the satellite?
vredina [299]

Answer:

-8.34\times 10^{11}\ \text{J}

Explanation:

m = Mass of satellite = 15000 kg

h = Distance above Earth = 800 km

R = Radius of Earth = 6371 km

M = Mass of Earth = 5.972\times 10^{24}\ \text{kg}

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Potential energy is given by

U=-\dfrac{GMm}{R+h}\\\Rightarrow U=-\dfrac{6.674\times 10^{-11}\times 5.972\times 10^{24}\times 15000}{(6371+800)\times 10^3}\\\Rightarrow U=-8.34\times 10^{11}\ \text{J}

The potential energy of the satellite is -8.34\times 10^{11}\ \text{J}.

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because it is not over the world please me 25 kilometre

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