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3 years ago

8

Consider a makeup mirror that produces a magnification of 1.5 when a person's face is 11.5 cm away what is the focal length of t

he makeup mirror in meters?

1 answer:

LiRa [457]3 years ago

4 0

<u>Answer:</u> The focal length of makeup mirror is 0.3448 m.

<u>Explanation:</u>

We are given:

m = 1.5

u = -11.5 cm

To calculate the image distance, we the equation for mirror:

$m=-\frac{v}{u}$

Putting values in above equation, we get:

$1.5=-(\frac{v}{-11.5})\\\\v=17.25cm$

To calculate the focal length of mirror, we use the mirror formula, which is:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Putting values in above equation, we get:

$\frac{1}{f}=\frac{1}{17.25}+\frac{1}{-11.5}\\\\f=-34.48cm$

Negative sign represents the position of focal length.

Converting the focal length into meters, we use the conversion factor:

1 m = 100 cm

So, 34.48 cm = 0.3448 m

Hence, the focal length of makeup mirror is 0.3448 m.

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Answer:

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Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

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$\bar{AB}=(h-0)i+(0-(-b))j$

$\bar{AB}=hi+bj$

We need to calculate the unit vector along AB

$u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}$

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$\hat{F}=F(u_{AB})$

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$\hat{M}=\hat{F}\times OA$

Put the value into the formula

$\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})$

$\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}$

Put the value into the formula

$\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}$

$\hat{M}=-81.102\ \hat{k}\ N-m$

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

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