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2 years ago

The power generated by an electrical circuit (in watts) as a function of its current xxx (in amperes) is modeled by:

Mathematics

2 answers:

vodka [1.7K]2 years ago

5 0

Answer:

300watts

Step-by-step explanation:

Given the power generated by an electrical device modeled by the equation

P(x) = -12x²+120x

The maximum power generated occurs when dP(x)/dx = 0

dP(x)/dx = -24x+120 = 0

-24x + 120 = 0

-24x = -120

x = 120/24

x = 5

Substitute x = 5 into the modeled function to get the maximum power as shown;

Recall that: P(x)=-12x^2+120x

P(5) = -12(5)² + 120(5)

P(5) = -12(25) + 600

P(5) = -300+600

P(5) = 300

Hence the maximum power generated by the circuit is 300watts

Irina18 [472]2 years ago

5 0

Answer:

300watts

Step-by-step explanation:

Given the power generated by an electrical device modeled by the equation

P(x) = -12x²+120x

The maximum power generated occurs when dP(x)/dx = 0

dP(x)/dx = -24x+120 = 0

-24x + 120 = 0

-24x = -120

x = 120/24

x = 5

Substitute x = 5 into the modeled function to get the maximum power as shown;

Recall that: P(x)=-12x^2+120x

P(5) = -12(5)² + 120(5)

P(5) = -12(25) + 600

P(5) = -300+600

P(5) = 300

Hence the maximum power generated by the circuit is 300watts

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3 years ago

Solve for x and state the measure of the angles with an x in them

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Answer:

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Step-by-step explanation:

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2 years ago

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Aleksandr [31]

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2 years ago

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Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

L = luminosity of the star (related to the Sun)

d = distance in ly (light-years)

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

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2 years ago