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pentagon [3]
3 years ago
6

A stone was dropped off a cliff and hit the ground with a speed of 136 ft/s. what is the height of the cliff? (use 32 ft/s2 for

the acceleration due to gravity.)
Physics
1 answer:
mina [271]3 years ago
7 0
<span>(v1)^2=(v0)^2-2ad; v1= final velocity; v0= initial velocity; a=gravitational constant;d=distance Since the stone was dropped we know v0=0. (v1)^2=2ad (136)^2=2(32)d d=(136)^2/(2*32) d=289ft</span>
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Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
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<h3>b.</h3>
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<h2>a.</h2>

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I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

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I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

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<h2>b</h2>

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od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

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od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

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od( 1\ cm) =  0.2606

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od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

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