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3 years ago

36

Mathematics

1 answer:

alexdok [17]3 years ago

3 0

Step-by-step explanation:

w(idth) = 2.5 inches for both boxes

h(eight) = 10 inches for both boxes

l(ength)l(arge) = 8 inches

l(ength)s(mall) = 75% of ll (=> of 8 inches)

=> ls = 8 × 75/100 = 2 × 75/25 = 2 × 3 = 6 inches

vl (volume large) = 8 × 2.5 × 10 = 200 inches³

vs (volume small) = 6 × 2.5 × 10 = 150 inches³

the difference of these two volumes is 50 inches³

the units to be used is cubic inches (inches³). we are talking about volumes. not lengths (inches), not areas (inches²).

the reason - for a volume we need to multiply 3 lengths (measured in simple inches). so, inch multiplied by inch is inch². this multiplied by inch again is then inch³.

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3 years ago

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]

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Step-by-step explanation:

The apparent brightness of a star is

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

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d = distance in ly (light-years)

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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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