At what distance is an object if its parallax as measured from either end is 1000km baseline (a)1 (b) 1' (C)1"
149112
Read the book and understand then you can figure it out yourself. that is the reason for school.
881212
Parallax is the method that uses triangulation. The diameter of the earth's orbit around the sun is 186,000,000 miles. A "nearby" star can be observed twice six months apart, and it will seem to have moved against the background of "distant" stars. Trigonometry can then be used to estimate the distance to the nearby star. \n\nThe particular months of observation depend on the location of the star. First, you have to know in which month the star appears due south at midnight ("in opposition"). Then you make your two observations three months before and three months after. Of course, you have to wait until the following year to make your "three months earlier" observation.\n\nYou can see the parallax effect in your front yard. Stand in your front yard so that a telephone pole on your side of the street is slightly to the right of a telephone pole across the street. Now move to the right. The nearby pole will move to the left of the distant one.
960920
0.07896 km\n153 mm\n0.578  km
749447
Hi. A good one might be Regulus because it lies pretty close to the ecliptic and is very bright. But you can use any star. For much better accuracy scientists use quasars due to their vast distance. This eliminates parallax error.
236037
Let's talk about parallax first - this is where parsec came from. Suppose that you live in a house on a beach and you see a beautiful island and the see and you would like to know, how far the island is. If you look at tiny object on the island (e.g., a light house) from your bedroom, which is on one end of the house and then you look at the same object from your kitchen, which is on the other end of the house, you'll notice that the viewing angle has changed slightly. If you measure this change of viewing angle and if you know how far you kitchen is from your bedroom, you'll be able to figure out the distance to the island without getting there. This is how astronomers measure the distance to stars (island) by using Earth's rotation around the Sun. The difference is that instead of walking from kitchen to the bedroom they have to wait for six months and the distance is the diameter of the Earth orbit. Reciprocal of one half of this change in angle is the parsec.
1017869
S=1.317h 1/2  S is the distance from to the horizon, h is how much above sea level. The AMS states that the horizon is at minimum 3 miles away.   So for a person whose eye level is 5 ft above sea level, the horizon is 1.317m(5ft)(1/2) which equals 3.29 miles
458976
The answers for these questions:\n\nAnswer 1: (c) 27 meters\nAnswer 2: (d) .5 km\n\nAlthough the answer to 2nd question should be 0.4 kms, but I reckon it's just approximate. \n\nYou can email me for further help by clicking Contact Nitro in my answers profile.
361687
Use the lense equation.\n\n1/f = 1/u + 1/v\n\nF is the focal length of the lense (1/f is the power of the lens as in your question)\nU & V are the near and far sighted distances.\n\nFor close up photography assume v= infinity (1/v = 0)\nU is the distance from your lens of the object being photographed (in metres).\n\nIe if you are photgraphic 20cm away, u=0.2. 1/u = 5 = the power of the lense you want.
894210
focal length is one half the radius of curvature, which is 22.5 cm.  the magnification is proportional to the image distance divided by the object distance.  So the image's distance is twice the object's distance from the mirror.  Now, 1/(object distance)+1/(image distance)=1/(focal length).  Or, 1/(2*object distance)+1/(object distance)=1/22.5 cm.  After some algebra, you can find that the object's distance is 3/2 times the focal length, which is 33.75 cm.  Check the math if you like.  You have all of the equations now.
374444
100\n\n(10mm=1cm; 10x10=100)
1000825
Sell the barometer and use the money to hire someone to find the answer.\n\nFind a point a which the barometer and the building have the same apparent height. Measure the distance from from eyeball to the barometer and the distance from your eyeball to the building. The ratio of the distances will equal the ratio of the heights. \n\nAt least I tried to give original answers!
764566
a) 5000*,4 = 2000 feet, so a = 1000 feet\n\nb) x^2/a2 -y2/b^2 =1, where;\n   where the two foci are located at (-c,0) and (c, 0) so c= 4000 ft\n   2a = difference in distances between foci and whale\n   b^2 = c^2-a^2\n\nc) Closest when y = 0 which give x= abs(a)\n\nd) y=+ or - b/a[sqrt ( x^2-a^2)], so if x < a, then y is imaginary. Therefore the region between x=a and x=-a must be excluded, that is, the whale can never be in that region. The whale will always be on the hyperbola branch that is associated with the mike that first receives the signal. The whale can be on either branch of the hyperbola, but never (obviously) on both at the same time. The distance from one foci to the whale will always be < or > than the distance from the other foci to the whale. The distances can never be equal. If they were, this technology would not work.\n\nI believe I included all the info you need to answer the question completely. It would be nice if a diagram could be included.
328904
2.54 cm\n1609.3440058 m\n39.37007874 in\n0.62137119 miles
989734
Two I have seen are AU where 1 AU or Astronomical Unit is the distance between the Earth and Sun, and LD which is the Lunar Distance, approx 250000 miles.\n\nLargest? Parsec: A unit of astronomical length based on the distance from Earth at which stellar parallax is one second of arc and equal to 3.258 light-years, 3.086 × 10^13 kilometers, or 1.918 × 10^13 miles.
620878
focal length = infinity\ncurvature  = 0
508808
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
869579
let the height of cliff be h\nthen \ndistance of first ship =x\ndistance of first ship =x+500\nnow \n1.h/x=tan 24\n2.h/(x+500)=tan 15\nfrom 1\nx=hcot24\nfrom 2\nx=hcot15-500\ntherefore\nh(cot15-cot24)=500\nh=336.47