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Naddik [55]
4 years ago
9

The densities of cardboard, aluminum, and lead are 0.6 g/cm^3, 2.7 g/cm^3, and 11.4 g/cm^3, respectively. Suppose that you are s

tudying the range of a (nonexistent) elementary particle, the Heidbrinkion, and that it takes 49 cm of cardboard, or 42 cm of aluminum, or 17 cm of lead to stop half of the Heibrinkions emitted from a source. Calculate the absorber thickness for each material.
__________ g/cm^2 for cardboard
__________ g/cm^2 for aluminum
__________ g/cm^2 for lead
Physics
1 answer:
krek1111 [17]4 years ago
4 0

Answer:

For cardboard = 29.4 g/cm²

For aluminium = 113.4 g/cm²

For lead = 193.8 g/cm²

Explanation:

Given:  

Density of the cardboard, d₁ = 0.6 g/cm³

Density of the aluminium, d₂ = 2.7 g/cm³

Density of the lead, d₃ = 11.4 g/cm³

Length of the cardboard,  L₁ = 49 cm

Length of the aluminium, L₂ = 42 cm

Length of the lead, L₃ = 17 cm

Now,

The absorber thickness is calculated as:

= Density × Length

therefore,

For cardboard = d₁ × L₁ = 0.6 × 49 = 29.4 g/cm²

For aluminium = d₂ × L₂ = 2.7 × 42 = 113.4 g/cm²

For lead = d₃ × L₃ = 11.4 × 17 = 193.8 g/cm²

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