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3 years ago

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The relationship between voltage, E, current, I, and resistance, Z, is given by the equation E = IZ. If a circuit has a current

I = 3 + 2i and a resistance Z = 2 – i, what is the voltage of the circuit?. a. 4 – i. b. 4 + i. c. 8 + i. d. 8 + 7i

2 answers:

Alex787 [66]3 years ago

7 0

By definition, we have to:

$E = IZ$

Where,

E: voltage

I: current

Z: Resistance

To calculate the voltage value, we must replace the following values in the given equation:

$I = 3 + 2i\\Z = 2 - i$

We have then:

$E = (3 + 2i) (2 - i)$

Making the product of complex numbers we have:

$E = (3 * 2 - 3i) + (4i - 2i ^ 2)\\E = (6 - 3i) + (4i - 2 (-1))\\E = (6 - 3i) + (4i +2)$

$E = (6 + 2) + (4-3) i\\E = 8 + i$

Answer:

the voltage of the circuit is:

$E = 8 + i$

option C

goldfiish [28.3K]3 years ago

3 0

The correct answer is C. 8 + i

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Read 2 more answers

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

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In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

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Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

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800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

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$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

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