1. Which 0f the following is equal to 4 kilograms? \n\nА. 4,000 cg \nB.40 g\nС. 4,000 mg \nD. 4,000 g\n\n2. What is the sum 0f 6 feet 10 inches and 8 fееt  9 iпсhеs? \n\nА. 14 ft. 7 in.\n В. 15ft. 7 in. \nС. 13 ft. 19 in. \nD. 15 ft. 19 in. \n\n3. What! is the elapsed time bеtwееn 3:40 А.М. апd  2:00 Р.М.? \n\nА. 8 hr. 20 miп. \nВ. 11 hr. \nС. 10 hr. 20 min. \nD. 1 hr. 40 miп. \n\n4. How many yards аге in 5 miles? \n\nА. 8,800 yd. \nВ. 5,280 yd. \nС. 1,760 yd. \nD. 26,400 yd.
203444
1. D. 4000 g. because there are 1000 grams in a kilogram.\n\n2. B. 15 ft. 7 in. because you just have to add the feet and then add the inches (14 ft. 19 inches) and then change twelve of the inches into one foot. \n\n3. C. 10 hr. 20 min. because it is 8 hr. 20 min. from 3:40 AM to noon, and it's 2 more hours to 2:00 PM. \n\n4. A. 8,800 yd. because there are 1,760 yards in one mile, and then you just multiply by 5. \n\nI hope this helps. Don't just copy the answers, but make sure you UNDERSTAND the material. :)
527006
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
199005
1.) A - 0.6666666\n2.) C - $12.26\n3.) B - 0.308\n4.) C - 19.87\n5.) A - $.65
256118
1) 4 3/4\n\n2) 1356.66\n\n3) 760
477846
a) 1500 * 2.54 = 3810 cm.\nb) -\nc) -\n\nOH MY GOD... I DONT KNOW..
504003
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!
514565
1. d=t, any form of d=t+b would be parallel. b = any number. That is a line.\n\n2. d=t would also work here.\n\n3. Well if you go 60 MPH and you wnat to know how many miles a minute you can go convert using (60miles/hour)*(1hour/60minutes) and you get 1mile/minute. If you need feet, 1 mile is 5280 feet, so 5280 feet/minute
234264
1(A) 1 L = 1000 cm^3 therefore 2L = 2000 cm^3\n1(B) 1 L = 1 Kg therefore 2L = 2Kg\n\n*******************\n\n2)\nTwo thirds less than half of what number is five sixths\n(1/2)X - (2/3) = (5/6)\n\nmultiply every term by 6\n3x - 4 = 5\n\nadd 4 to both sides\n3x = 9\nx = 3\n\n*******************\n\n3)\nvolume of a cube = side^3 = 12^3 = 1728 cubic inches\nvolume of square pyramid = (1/3)(b^2)(h)\n where b = 12 and h = 8 (because the total height = 20)\n\n v = (1/3)(144)(8) = 384 cubic inches\n\n total volume = 2112 cubic inches\n\n*******************\n\n4)\nwe do not know the relation ship between the segments, if they form a line, if they form a right triangle, or if they lie on top of each other....\n\nif they form a line segment, where "b" is between "a" and d", then the total meximum length is 12 + c. if they lie on top of each, then they have a minimum distance of |12 - c| (absolute value of 12 - "c")\n\n*******************\n\n5)\nyou would need (1 x 10^3 or 1000) to form 1 meter multiplied by 1000 to form 1 KM\n1000 x 1000 = 1,000,000 or 1 x 10^ 6\n\n*****************\n6)\nx^2 + x  : GCF = x\nx(x + 1)\n\n\n12m^2n^3 + 18mn^2 - 24m^2n^3 \n12m^2n^3 and 24m^2n^3 are like terms\ntherefore\n\n= -12m^2n^3 + 18mn^2\n\nGCF = 6mn^2\n\n6mn^2(-2mn + 3) or 6mn^2(3 - 2mn)\n\n*****************\n\n7)\n-2 2/3w - 1 1/3 = 4 \n\nadd 1 1/3 to both sides\n\n-2 2/3w = 5 1/3\n\n= -(8/3)w = (16/3)\n\nmultiply both sides by -(3/8)\n\nw = -2
383019
1)\nF = mg = kx\n8 * 9.8 = k * 0.1\nk = 784 N/m\n\n2)\nelastic PE = (1/2) * kx^2\nelastic PE = (1/2) * 784 * 0.4^2\nelastic PE = 62.72 N-m\n\n3)\nAt maximum height, all elastic PE stored in the spring becomes the gravitational PE in the stone.  Therefore, the gravitational PE equals...\n62.72 N-m \n\n4)\nelastic PE = gravitational PE = mgh\n62.72 = 8 * 9.8 * h\nh = 0.8 m = 80 cm (from release pt)
396611
F=F1-F2\nF=qm(M1/R1^2-M2/R2^2)\nWhere \nq- universal gravitational constant (6.672 x 10^-11 N m^2/kg^2) \nM1- mass of  407kg\nM2 – mass of  102kg\nM - 21.7 kg\nR1-distance between M1 and the object\nR1-distance between M2 and the object\n\nIn the first case R1=R2=Rso we have \n\nF=qm(M1 -M2)/R^2\nF=6.672 x 10^-11 x 21.7(407 + 102kg) / (0.475)^2=3.266e-6 N\n\n\n\nFor the second question \n\nF=F1-F2=0\n\nSo \n\nM1/R1^2=M2/R2^2\n\nR1/R2 = (M2/M1)^.5\n\nAnd we know that R1+R2=0.475 m\nSo R2=0.475 –R1\n\nThen we have \n\nR1 = (M2/M1)^.5 (0.475 –R1)\nR1=((M2/M1)^.5  x 0.475)/(1+(M2/M1)^.5)\nR1=(1.997 x.475)/(1 + 1.997)= 0.317 m\nAnd R2 must me then 0.475-.317=0.158  m
427432
k, your questuons are hard to understand but I'll try and help\n\nUnit conversions are simple, you write them out like fractions, impossible on a computer screen normally...but maybe like this\n\nExample: convert 12 ft/sec into meters.\n\n(12 feet)        (1 meter)\n------------    ---------      = 3.657  meters per sec\n(second)       (3.28 feet)\n\nthe feet cancel\n\nSo basically you multiplied it by 1, because the numerator and denominator are the same thing, 12 times 1, but with different units.
497886
1) \n76>40+4c\n36>4c\nc<9\n\n2) 1500>100+17.5p\n1400>17.5p\np<80
145556
a) d= 0.5 * 9.8 * t^2 + Vi * t = 10 m\nFor t=1 sec:\nVi= 10 - 9.8*0.5 = 5.1 m/s \n\nb) g= (Vf-Vi)/ t\nFor t=1sec and Vi=5.1\nVf= 9.8+5.1= 14.9 m/s\n\nHope it is clear enough.
234105
A) F = mg = 50 kg x 9,81 m/s2 = 0,49 kN.\nB) F = GMm/R2 = 6.67E-11 x 5.98E24 x 50 / (6.370E6)^2 N\n   = 0,49 N
404410
F=-kx and F=ma\n\nfind mass from period     T = 2pi(m/k)^(1/2)\n\nhmmmm,  you seem to be missing k\n\nfor fixed ratio m/k the period of 0.51s and amplitude 4 cm is not enough to determine the mass (and therefore acceleration).
538387
1) two hours from 9 to 11. so distance 45*2 = 90 km\n\n2) The car has done 100+108= 208 km in 5 hours\nthe speed is 208/5 = 41.6 kph
463159
1. b\nBecause \n'1/s' is inversly related to 's'\n\n2. c\nBecause\n1/s AKA Hertz\n\n3. d\nBecause\nN.P,As,Bi are all in the same chemical group\n\n4. b\nBecause\n(3*10^8)m/s / (2.50 * 10^13)Hz = 1.20 x 10^-5 m\n\n5. b\n\n(I am 100% sure for every question accept the last one. I am only 50% sure on 5)