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e-lub [12.9K]
3 years ago
10

A statue of a great baseball player weighs 2400 N and has a base that is 4 m x 2 m. What is the pressure the statue exerts on th

e floor?
Physics
1 answer:
bulgar [2K]3 years ago
5 0

Answer:

300N/m²

Explanation:

Pressure is calculated using the formula;

Pressure = Force/Area

Given

Force = 2400N

Area = 4m×2m = 8m²

Substitute the given parameters into the formula as shown;

Pressure = 2400/8

Pressure = 300N/m²

Hence the pressure the statue exerts on the floor is 300N/m²

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I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
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Answer:

<h3>a.</h3>
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  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
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Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

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}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

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

I(2 \ cm) = I_0 e^{- 1.2}

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<h2>b</h2>

The optical density od is given by:

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

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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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