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gogolik [260]
2 years ago
11

Many physical quantities are connected by inverse square laws, that is, by power functions of the form f(x)=kx^(-2). In particul

ar, the illumination of an object by a light source is inversely proportional to the square of the distance from the source. Suppose that after dark you are in a room with just one lamp and you are trying to read a book. The light is too dim and so you move halfway to the lamp. How much brighter is the light?
Physics
1 answer:
goldenfox [79]2 years ago
3 0

Answer:

  • 4 times

Explanation:

Since the equation for the illumination of an object, i.e. the brightness of the light, is <em>inversely proportional to the square of the distance from the light source</em>, the form of the function is:

  • f(x) = k.x⁻²

Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.

Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :

F=k(x/2)^{-2}=\frac{k}{(x/2)^2}= \frac{k}{x^2}. 4=f(x).4

Then, you have found that the light is 4 times as bright as it originally was.

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\sf{Here \:  Resistances  \: are ,}

\sf• \:  R_{1}  =3  Ω

\sf•  \: R_{2}  =3Ω

\sf•  \: R_{3}  =3Ω

\sf{We \:  know  \: the  \: formula  \: of  \: the  \: Equivalent \:  Resistance \:  for  \: Series  \: Circuit, }

<h3>\bf \purple {\bigstar  {\: R_{s}  =  R _{1}  +  R _{2}  +  R_{3}+...+ R_{n}}}</h3>

\sf ⇒ R_{s} =(3 + 3 + 3)Ω

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\sf  \pink{ \boxed{Answer : 9 Ω.}}

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\sf{(ii)  \: We \:  are \:  given  \: a  \: figure \:  of  \: Parallel \:  Circuit. }

\sf{Here \:  Resistances  \: are ,}

\sf• \:  R_{1}  =3  Ω

\sf• \:  R_{2}  =3  Ω

\sf• \:  R_{3}  =3  Ω

\sf{We \:  know  \: the  \: formula  \: of  \: the  \: Equivalent \:  Resistance \:  for  \: Parallel  \: Circuit, }

<h3>\bf \purple {\bigstar  {\:  \frac{1}{R_{p}}  =   \frac{1}{R _{1}}  +   \frac{1}{R_{2}}  +  \frac{1}{R_{3}}+...+\frac{1}{R_{n}} }}</h3>

\sf⇒ \frac{1}{ R_{p} }  =  \frac{1}{3}  +  \frac{1}{3}  +  \frac{1}{3}

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\sf⇒ \frac{1}{ R_{p} }  =  \frac{3}{3}

\sf⇒ \frac{1}{ R_{p} }  =  1

\sf \therefore R_{p}   =  1Ω

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5 0
3 years ago
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In 1984 a team of german physicists synthesized a new element by smashing an iron-58 nucleus into a lead-208 target. if a neutro
dimulka [17.4K]

Here in all nuclear reactions we can say that mass conservation and charge conservation is always true

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58 + 208 = x + 1

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now we will use charge conservation

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3 years ago
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Answer:

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