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MakcuM [25]
3 years ago
12

A cube has a drag coefficient of 0.8. what would be the terminal velocity of a sugar cube 1 cm on a side in air (ρ = 1.2 kg/m3)?

take the density of sugar to be 1.6 x 103 kg/m3.
Physics
1 answer:
kramer3 years ago
7 0
To answer this question, we should know the formula for the terminal velocity. The formula is written below:

v = √(2mg/ρAC)
where
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient

Let's determine the mass, m, to be density*volume.
Volume = s³ = (1 cm*1 m/100 cm)³ = 10⁻⁶ m³
m = (1.6×10³ kg/m³)(10⁻⁶ m³) = 1.6×10⁻³ kg
A = (1 cm * 1 m/100 cm)² = 10⁻⁴ m²

v = √(2*1.6×10⁻³ kg*9.81 m/s²/1.6×10³ kg/m³*10⁻⁴ m²*0.8)
<em>v = 0.495 m/s</em>
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Answer:

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Heat gained by the water:

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q = (70 g) (4.2 J/g/K) (40°C − 0°C)

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q = mL + mCΔT

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q = 4880 J

Heat lost by the steam:

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q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)

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m = 7.6 g

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