Mastering Physics chapter 6 -6.65 Large objects have inertia and tend to keep mo
ID: 1278378 • Letter: M
Question
Mastering Physics chapter 6 -6.65
Large objects have inertia and tend to keep moving-Newton's first law. Life is very different for small microorganisms that swim through water. For them, drag forces are so large that they instantly stop, without coasting, if they cease their swimming motion. To swim at constant speed, they must exert a constant propulsion force by rotating corkscrew-like flagella or beating hair-like cilia. The quadratic model of drag of the equation given below fails for very small particles. Instead, a small object moving in a liquid experiences a linear drag force, D? = (bv, direction opposite the motion), where b is a constant. For a sphere of radius R, the drag constant can be shown to be b=6??R, where ? is the viscosity of the liquid. Water at 20?C has viscosity 1.010?3N?s/m2.
D? = (12CAv2 , direction opposite the motion).
A paramecium is about 100 ?m long.
Express your answer to two significant figures and include the appropriate units.
1. If it's modeled as a sphere, how much propulsion force must it exert to swim at a typical speed of 0.70mm/s ?
2. How about the propulsion force of a 2.0-?m-diameter E.coli bacterium swimming at 40?m/s ?
3. The propulsion forces are very small, but so are the organisms. To judge whether the propulsion force is large or small relative to the organism, compute the acceleration that the propulsion force could give each organism if there were no drag. The density of both organisms is the same as that of water, 1000 kg/m3.
4. Compute the acceleration that the propulsion force could give E.coli bacterium if there were no drag.
Please show work with answers
I have a really bad grade so far and need to get 100% of the points to do well on this assignment. So I need the answers as well as how to work out the question.
Explanation / Answer
Answer for A: 100 micrometer swimming at 1mm/s
A: 9.4 x 10^(-10) N use D=bv (this is given) b=6(pi)(n)R where R = 100 micro meters (100/1000000)/2 and n= .001 N s/m^2 (this is given)
6(pi)(.001)(.00005) = b then b(v) = b(.001 m/s) (v is given a 1 mm/s convert to m/s)
Answered B for 2.0 micrometer swimming at 22 micrometers/s
B: 4.1 x 1-^(-13) N same formula as in A but with different R and v
C: 1.8 m/s^2 V=(4/3)(pi)(r^3), m = V(density) = V(1000 kg/m^3) now you have mass
F=ma so a=F/m use your F rom part A and the mass you just calculated
D: 99 m/s^2 same as part C but different R and F(use the ones from part B)
You have a lot of really small numbers to work with so pay careful attention