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Setler [38]
3 years ago
11

How to solve this question

Engineering
1 answer:
Harrizon [31]3 years ago
3 0

Answer:

Voltage across element X is 49.26 volts

Explanation:

The kirchhoff's voltage law states that the algebraic sum of voltages around a closed loop is equal to zero.

Applying the kirchhoff's voltage law in loop yields,

-160 + V₁ + V₂ + V₃ + Vₓ + 60 + V₄ = 0

-160 + 60 + V₁ + V₂ + V₃ + Vₓ + V₄ = 0

But we know that V = IR

-100 + IR₁ + IR₂ + IR₃ + IRₓ + IR₄ = 0

IR₁ + IR₂ + IR₃ + IRₓ + IR₄ = 100

I(R₁ + R₂ + R₃ + Rₓ + R₄) = 100

I = 100/(R₁ + R₂ + R₃ + Rₓ + R₄)

Substitute the given resistor values

I = 100/(18k + 20k + 45k + 100k + 20k)

I = 100/203k

I = 100/203×10³

I = 0.0004926 A

The voltage drop at element x is

Vₓ = IRₓ

Vₓ = 0.0004926*100×10³

Vₓ = 49.26 V

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(i) what assumptions about the relationship between the inputs and output are inherent in this specification? do scatter plots s
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A problem that will be handled by a procedure is described by an input-output specification.

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Please help, Artificial Intelligence class test
MrRissso [65]

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4 0
2 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
GenaCL600 [577]

Answer:

62.14\ \text{miles}

6213727.37\ \text{miles}

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = 10^5\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}

1\ \text{mile}=1609.34\ \text{m}

\dfrac{10^5}{1609.34}=62.14\ \text{miles}

The chain would extend 62.14\ \text{miles}

Dislocation density = 10^{10}\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}

\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}

The chain would extend 6213727.37\ \text{miles}

3 0
3 years ago
A train consists of a 50 Mg engine and three cars, each having a mass of 30 Mg . If it takes 75 s for the train to increase its
ohaa [14]

Answer:

T = 15 kN

F = 23.33 kN

Explanation:

Given the data in the question,

We apply the impulse momentum principle on the total system,

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

we substitute

[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F( 75 - 0 ) =  1.75 × 10⁶

The resultant frictional tractive force F is will then be;

F =  1.75 × 10⁶ / 75

F = 23333.33 N

F = 23.33 kN

Applying the impulse momentum principle on the three cars;

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F(75-0) = 1.125 × 10⁶

The force T developed is then;

T =  1.125 × 10⁶ / 75

T = 15000 N

T = 15 kN

7 0
2 years ago
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