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Zigmanuir [339]

2 years ago

14

A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location

of the image?A)6.15 cm to the left of the lensB)26.7 cm to the right of the lensC)6.15 cm to the right of the lensD)6.00 cm to the right of the lensE)26.7 cm to the left of the lens

1 answer:

amm18122 years ago

6 0

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

$f=-16.0 cm$ (the focal length is negative for a diverging lens)

$p=10.0 cm$ is the distance of the object from the lens

Solvign the equation for q, we find

$\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}$

$q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm$

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

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2 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

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Explanation:

Given that,

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$f_{c}=\dfrac{1}{2\pi R C}$

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$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

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Using formula of $V_{R}$

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$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

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(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

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Using formula of $V_{R}$

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$V_{R}=4.561\ Volt$

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Answer:

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Answer:

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