Steel baĺl of 3mm radius is falling in glycerine find terminal velocity
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we know that terminal velocity formula is:F=6πvrηmg =6πvrη6πη = mg/vrmg is the weight of one drop of radius 1 mmIf they combine :10mg = 6πv'r'η10*4/3*π*r3 = 4/3*π*r'3r' = 2.1544 r10 mg = 6πv'r'η10mg = v'r'*mg/vr10 vr = v'r'10vr = 2.1544 r* v'v' = 4.64 v
we know that terminal velocity formula is:F=6πvrηmg =6πvrη6πη = mg/vrmg is the weight of one drop of radius 1 mmIf they combine :10mg = 6πv'r'η10*4/3*π*r3 = 4/3*π*r'3r' = 2.1544 r10 mg = 6πv'r'η10mg = v'r'*mg/vr10 vr = v'r'10vr = 2.1544 r* v'v' = 4.64 v
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we know that terminal velocity formula is:F=6πvrηmg =6πvrη6πη = mg/vrmg is the weight of one drop of radius 1 mmIf they combine :10mg = 6πv'r'η10*4/3*π*r3 = 4/3*π*r'3r' = 2.1544 r10 mg = 6πv'r'η10mg = v'r'*mg/vr10 vr = v'r'10vr = 2.1544 r* v'v' = 4.64 v
Explanation:
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