saplinglearning.com University-PHYS 211-Fall17-STINNETT Activities and Due Dates
ID: 1791892 • Letter: S
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saplinglearning.com University-PHYS 211-Fall17-STINNETT Activities and Due Dates > HW: Momentum Impulse and Collisions 11/10/2017 10:00 PM 83.3/100 O 10/31/2017 12:33 AM -Print calculator-dPenodic Table Question 19 of 19 Sapling Learning Map The figures show a hypothetical bodles are given in Astronomical system is zero. Find the magnitude, ds, of the star's displacement system at two different times. The spatial coordinates (x, y) of the Units (AU). In the first picture, the velocity of the center of mass of ms 2.5085 x 100 kg mA 2.8067x102kg AUI mb = 6.8307 x10as kg mc 8.2507 x10 kg Number (0, 1.6711) (0.9349, 1.2553) (0.1119, 0) (-1.5385, 0) (0, 0) (0, -0.2803) (-0.5911, -0.5553) O Previous Check Answer Next Exit intExplanation / Answer
from the given data
ms = 2.5085*10^30 = 25085k
ma = 2.8067*10^28 = 280.67k
mb = 6.8307*10^26 = 6.8307k
mc = 82.507*10^26 = 82.507k
where k = 10^26
so the center of mass of this system will not change its locaiton
let its location be (X,Y)
now, initially
(xs,ys) = (0,0)
(xa,ya) = (0.1119,0)
(xb,yb) = (0.9349,1.2553)
(xc,yc) = (0,1.6711)
hence
(X,Y) = (msxs + maxa + mbxb + mcxc, msya + maya + mbyb + mcyc)/(ms + ma + mb + mc)
finally let position of sun be (x,y)
(x's,y's) = (x,y)
(x'a,y'a) = (0,-0.2803)
(x'b,y'b) = (-1.5385,0)
(x'c,y'c) = (-0.5911, -0.5553)
hence
(X,Y) = (msx's + max'a + mbx'b + mcx'c, msy'a + may'a + mby'b + mcy'c)/(ms + ma + mb + mc)
as (X,Y) remains the same
(msxs + maxa + mbxb + mcxc, msya + maya + mbyb + mcyc)/(ms + ma + mb + mc) = (msx's + max'a + mbx'b + mcx'c, msy'a + may'a + mby'b + mcy'c)/(ms + ma + mb + mc)
(maxa + mbxb , mbyb + mcyc) = (msx + mbx'b + mcx'c, msy + may'a+ mcy'c)
hence (x,y) = (0.003869719, 0.01080087) AU
so displacement of star = sqroot(x^2 + y^2) = 0.011473 AU
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