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kiyahlove7537
27.02.2020 •
Physics
A sphere made of rubber has a density of 1.00 g/cm3 and a radius of 8.00 cm. It falls through air of density 1.20 kg/m3 and has a drag coefficient of 0.500. What is its terminal speed (in m/s)
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Ответ:
Explanation:
m = p_sphere*V A = pi*r^2 v_t= sqrt(2mg/Dp_airA)
The Attempt at a Solution
m = (870 kg/m^3)(4/3)pi(0.085 m)^3 = 2.24 kg
A = pi(0.085 m)^2 = 0.0227 m^2 v_t = sqrt[2(2.24 kg)(9.80 m/s^2)/(0.500)(1.20 kg/m^3)(0.0227 m^3)] = 56.8 m/s
Ответ:
Terminal speed= 1,826.51m/s
Explanation:
Volume of a sphere is given by: V=4/3pir^3
Where r is radius of sphere
V=4×3.142×(8)^2/3
V= 2144.66cm^3
Converting to meters
V=2144.66cm^3×(1m^3/ 1×10^-6cm^3)
V= 2.145×10^-3m^3
Area of sphereA= pi(8)^2
A= 3.142×64=210.6cm^3
Converting to meter
201cm^×(1m/10000cm^2)
A=0.0210m^2
Given:
Density of shere= 1.00kg/m^3
Drag coefficient =0.500
Mass of sphere=?
Density of sphere= mass of sphere / volume of shere
Mass= 2144.66cm^3×1.00kgcm^3
Mass= 2144.66kg
Terminal speed,VT= Sqrt(2mg)/(DpA)
VT= Sqrt( 2×( 2144.66)×9.8))/(0.500×1.20×0.021)
VT= Sqrt(42035.34/0.0126)
VT=Sqrt(3,336,137.78)
VT= 1,826.51m/s
Ответ:
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