A solid aluminum sphere is dropped from an airplane and attains a terminal speed of 191.6 m/sIf the drag coefficient for the sphere was 0.476, what is the total volume of the sphere? _m3What is the mass of the sphere? _kgWe have hints:- P Al 2.70 x 10^3 kg/m3- P air 1.28 kg/m3Please show how to work this outAs I've already quot;taken a stab at this,quot; I like to compare the work I've done with yours to make sure I'm not missing any steps or better yet, to learn any steps I did miss.I will post quot;Best Answersquot; as Yahoo Allows(Sometimes even though I chose a quot;Best Answerquot; the community best answer shows up instead?) Thank you for your help!
Find the two ends that are loose and causing the hole and tie them so that they don't unravel moreDo not cut any extra threadI would take it to someone who crochets to fix itSomeone experienced will know how to do thatYou could also get a darning needle and try to weave the ends back in and make it look as good as you canAll of those little blocks are sewed together and there are a lot of tie-offs due to the color changesI consider your blanket to be delicateI would soak it in woolite to wash it, gently squeeze to wring out water, (no twisting), then roll up in a towel to get more water out of it, finally, spread out on a big towel to air dryNo dryersI would wash it only when it needed it.
I don't have the Crochet me book so without a picture I can only guessYou could try weaving a running stitch through some of the rows and gather it as tight as you want it to beIt will change the look of the stitching but save your workThen I would suggest lining itDon't feel too bad, I'm sure your Mom loved that you made it for herWe all have projects that.well, lets just say they taught us somethingLOL.
If the handle is that long can you pull it up at the shoulder and tie a knot? Alternatively stitch the handle to the right length with a sewing machine and cut off the excess.
Hi there Carolina, This is a fairly simple plug and chug problem as long as you have the right formulasFor a spherical object in free fall, the terminal velocity is: V_t sqrt( 2 m g / (density_fluid projected_area C_d)) M mass g gravity C_d drag coefficient The problem statement gives us everything but the projected area and the mass of the sphere Projected_area: Because the object is spherical the projected area is always going to be the same and it will be equal to the circular area around the center, picture staring straight up at a falling ball- no matter how it rotates the shape doesn’t change Projected_area (pi)radius_sphere^2 Mass: Because you have the density of aluminum, you can calculate the mass of the sphere using the volume Mass density_aluminum (4/3)(pi)radius_sphere^3 Now you can plug these two into the equation above for terminal velocity, and your only unknown will be the radius of the sphere! V_t sqrt( 2 [density_aluminum (4/3)(pi)radius_sphere^3] g / (density_fluid [(pi)radius_sphere^2] C_d)) Plugging in your known values the equation becomes: 191.6 m/s sqrt( 2[2700 kg/m3 4/3 3.14 radius)_sphere^3] 9.81 m/s2 / (1.28 kg/m3 [3.14 radius_sphere^2] 0.476]) Cleaning this up a bit: 191.6 m/s sqrt(221784.5radius_sphere^3 / 1.91 radius_sphere^2) 191.6 m/s sqrt(116117.53 radius_sphere) From this point you can easily solve for the radius Radius_sphere 191.6^2 / 116117.53 .316m Now you can go back to your equations for volume and mass and solve: V (4/3) pi radius_sphere^3 (4/3)3.14.316^3 .132m^3 M V density .132m^3 2700 kg/m3 356.7kg Hope this helps!
