Human blood contains plasma, platelets, and blood cells. Toseparate the plasma f

Question

Human blood contains plasma, platelets, and blood cells. Toseparate the plasma from other components, centrifugation is used.Effective centrifugation requires subjecting blood to anacceleration of 2000g or more. In this situation, assumethat blood is contained in test tubes of length L =15.7 cm that are full of blood. Thesetubes ride in the centrifuge tilted at an angle of 45.0° abovethe horizontal (a.) What is the distance (in cm) of a sample ofblood from the rotation axis of a centrifuge rotating at afrequency f = 3710 rpm, if ithas an acceleration of 2000g? (b) If the blood at the center of the tubes revolves aroundthe rotation axis at the radius calculated in Part (a), calculatethe minimum and maximum accelerations experienced by the blood ateach end of the test tube. Express all accelerations as multiplesof g. Human blood contains plasma, platelets, and blood cells. Toseparate the plasma from other components, centrifugation is used.Effective centrifugation requires subjecting blood to anacceleration of 2000g or more. In this situation, assumethat blood is contained in test tubes of length L =15.7 cm that are full of blood. Thesetubes ride in the centrifuge tilted at an angle of 45.0° abovethe horizontal (a.) What is the distance (in cm) of a sample ofblood from the rotation axis of a centrifuge rotating at afrequency f = 3710 rpm, if ithas an acceleration of 2000g? (b) If the blood at the center of the tubes revolves aroundthe rotation axis at the radius calculated in Part (a), calculatethe minimum and maximum accelerations experienced by the blood ateach end of the test tube. Express all accelerations as multiplesof g.

Explanation / Answer

then angular frequency is =(2/60)(3710)rad/s          = 388.5 rad/s we have centripetal acceleration a = r2 then r =(2000*9.8m/s2)/(388.5rad/s)2    = 0.1298m    = 13cm     b) Considering the the center of the tube is at r =0.13m then the lower part is at a distance of 0.13m +(0.157m/2)cos45o = 0.1855m then a = (0.1855m)(388.5rad/s2)    =27997.9m/s2   = (27997.9m/s2)/(9.8m/s2)g    = 2857g
and the top part is at a distance of 0.13m -(0.157m/2)cos45o = 0.07449m then a = (0.07449m)(388.5rad/s2)    =11242.9m/s2   = (11242.9m/s2)/(9.8m/s2)g    = 1147g
therefore the range of the acceleration is 1147g to2857g and the top part is at a distance of 0.13m -(0.157m/2)cos45o = 0.07449m then a = (0.07449m)(388.5rad/s2)    =11242.9m/s2   = (11242.9m/s2)/(9.8m/s2)g    = 1147g
therefore the range of the acceleration is 1147g to2857g

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