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3 years ago

1. calculate the speed of a wave that travels 10 cm in 4 seconds?

1 answer:

6 0

Use this equation
Distance(s)=speed(v)•time(t)
Given,req-soln
1.
S=10cm=0.1m
t=4sec
V=s/t=>0.1•10/4•10=>1/40=0.025
2.
V=0.1m/s
t=1.5 sec
S=vt =0.1•1.5=0.15

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Explanation:

Given that,

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(A). We need to calculate the crossover frequency

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

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C = capacitor

Put the value into the formula

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

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Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

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

$V_{R}=2.280\ Volt$

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

Using formula of $V_{R}$

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

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(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

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