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Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it holds 400.0 g of coffee when filled to a depth of 6.00 cm? Assume coffee has the same density of water. Answer in cm.

Respuesta :

Coffee mug is cylindrical shape. Therefore, 
Volume of mug = volume of cylinder = πr²h 

r = inner radius of mug. h = height of coffee = 6 cm. 
Density of coffee = Density of water = 1 g/ml 

Therefore, volume of coffee = mass/density = 400/1 = 400 ml = 400 cm³
Volume of coffee = πr²h
400= πr²(6) 
r² = 21.23
r = inner radius of mug = 4.607 cm
onyebuchinnaji

The inside radius of the coffee mug that can hold the given mass is 4.61 cm.

The given parameters;

  • mass of the coffee, m = 400 g
  • depth of mug, h = 6 cm
  • density of the coffee, ρ = 1 g/cm³

The volume of the coffee in the mug is calculated as follows;

[tex]volume = \frac{mass}{density} \\\\volume = \frac{400}{1} \\\\volume = 400 \ cm^3[/tex]

The inside radius of the coffee mug is calculated as follows;

[tex]Volume = \pi r^2 h\\\\400 = \pi \times r^2 \times 6\\\\r^2 = \frac{400}{6\pi} \\\\r^2 = 21.22\\\\r = \sqrt{21.22} \\\\r = 4.6 1 \ cm[/tex]

Thus, the inside radius of the coffee mug that can hold the given mass is 4.61 cm.

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