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alexira [117]
3 years ago
8

Consider a circuit element, with terminals a and b, that has vab= -12V and iab= 3A. Over a period of 2 seconds, how much charge

moves through the element? If electrons carry the charge, which terminal do they enter? How much energy is transferred? Is it delivered to the element or taken from it?
Engineering
1 answer:
miv72 [106K]3 years ago
5 0

Answer:

a) 6 coulombs

b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current

c) = -72 joules

Energy is taken from element

Explanation:

Given data:

V ab = -12 v

I ab = 3A

period ( t ) = 2 seconds

a) determine how much charge moves through the element

q = I * t

  = 3 * 2 = 6 coulombs

b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current

c) determine how much energy is transferred

= Vab * Iab * t

= -12 * 3 * 2

= -72 joules

Energy is taken from element

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3 years ago
**Please Help. ASAP**
natima [27]

Answer:

The answer is below

Explanation:

1)

\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at

2)

\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)

3)

x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}

4)

x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}

5)

x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}

6)

E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\  \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}

7)

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\  \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}

8)

ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\

4 0
3 years ago
What is the missing number in the pattern below?<br> 0, 1, 2, 3, 6, 11, 20, _?_, 68
lbvjy [14]

Answer:

0, 1, 2, 3, 6, 11, 20, 37, 68

Explanation:

Each number in the series, starting with 3, is the total of the three numbers before it. The following number is the total of the preceding three numbers, according to the pattern. The pattern's next number is 125.

3 0
3 years ago
At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge
tangare [24]

Answer:

e= 50 J/kg

Explanation:

Given that

Speed ,v= 10 m/s

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Density of the air ,ρ = 1.25 kg/m³

We know that mechanical energy given as

E=\dfrac{1}{2}mv^2\ J

That is why mechanical energy per unit mass will be

e=\dfrac{1}{2}v^2\ J/kg

Now by putting the values in the above equation we get

e=\dfrac{1}{2}\times 10^2\ J/kg

e= 50 J/kg

That why the mechanical energy unit mass will be 50 J/kg.

5 0
3 years ago
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