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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)

Respuesta :

Ayesha225

Answer:

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

anniwuanyanwu1

Answer:

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

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