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3 years ago

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James is planning on registering for a course in electrical engineering which of the following sub disciplines could he opt for

instrumentation engineering prosthetics plant design or signal processing

1 answer:

larisa [96]3 years ago

7 0

Answer:

Instrmentation engineering

Explanation:

Hope this helps!

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An antelope is being chased across the savannah by a hungry lion. The lion runs at a

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Answer:It can run in the same direction but it increases its speed.

Explanation:

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3 years ago

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In the 1660s, who discovered that color is a direct function of light by passing sunlight through a prism and observing the band

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<span>Sir Isaac Newton was the one who discovered that color is a </span><span>direct function of light by passing sunlight through a prism and observing the bands of the spectrum of colors through his newton of laws.</span>

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4 years ago

How much power does an electric device use if the current is 36.0 amps and the resistance is 3.9 ohms?

dsp73

<span>Question: How much power does an electric device use if the current is 36.0 amps and the resistance is 3.9 ohms? </span>

How?:

Equation: P = I^2 R

Meanings:
P = Power in Watts
I = Current in Ampere
R = Resistance in ohms.

Plugged in: P = 36^2<span> x 3.9 = 5054.4

Answer: P= </span>5100 watts.

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3 years ago

Can you answer this please i’ve been stuck on it for a while

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4 years ago

Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10^4 Pa.

Triss [41]

Answer:

9.8 × 10⁴Pa

Explanation:

Given:

Velocity V₁ = 12m/s

Pressure P₁ = 3 × 10⁴ Pa

From continuity equation we have

ρA₁V₁ = ρA₂V₂

A₁V₁ = A₂V₂

making V₂ the subject of the equation;

$V_{2} = \frac{A_{1}V_{1}}{A_{2}}$

the pipe is widened to twice its original radius,

r₂ = 2r₁

then the cross-sectional area A₂ = 4A₁

⇒ $V_{2}= \frac{A_{1}V_{1}}{4A_{1}}$

$V_{2}= \frac{V_{1}}{4}$

This implies that the water speed will drop by a factor of $\frac{1}{4}$ because of the increase the pipe cross-sectional area.

The Bernoulli Equation;

Energy per unit volume before = Energy per unit volume after

p₁ + $\frac{1}{2}$ ρV₁² + ρgh₁ = p₂ + $\frac{1}{2}$ ρV₂² + ρgh₂

Total pressure is constant and $P_{T}$ = P = $\frac{1}{2}$ ρV₂²ρV²

p₁ + $\frac{1}{2}$ ρV₁² = p₂ + $\frac{1}{2}$ ρV₂²

Making p₂ the subject of the equation above;

p₂ = p₁ + $\frac{1}{2}$ ρV₁² - $\frac{1}{2}$ ρV₂²

But $V_{2} = \frac{V_{1}}{4}$ so,

p₂ = p₁ + $\frac{1}{2}$ ρV₁² - $\frac{1}{2}$ ρ $\frac{V_{1}^{2}}{4^{2}}$

p₂ = 3.0 x 10⁴ + ( $\frac{1}{2}$ × 1000 × 12²) - ( $\frac{1}{2}$ × 1000 × 12²/4² )

P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³

P₂ = 9.79 × 10⁴Pa

P₂ = 9.8 × 10⁴Pa

4 0

3 years ago