How big is a 7.0 earthqauke?
476533
Earthquake Severity\n\nRichter         Earthquake\nMagnitudes      Effects\n\nLess than 3.5   Generally not felt, but recorded.\n\n3.5-5.4         Often felt, but rarely causes damage.\n\nUnder 6.0       At most slight damage to well-designed buildings.\nCan cause major damage to poorly constructed buildings\nover small regions.\n\n6.1-6.9         Can be destructive in areas up to about 100 kilometers\nacross where people live.\n\n7.0-7.9         Major earthquake. Can cause serious damage over larger areas.\n\n8 or greater    Great earthquake. Can cause serious damage in areas several\nhundred kilometers across.
853272
Use this online converter\nhttp://www.kt.rim.or.jp/~tyamamot/pt.htm\n\nPut in size then choose unit as mm\nConversion standard is American Standards Association 1933\nNumerical format is "real"\nClick "CALC"\nThen scroll down to "results" & look for the result for the 1/72 inch point system\nCross check: 1pt  (1/72inch point system) = .35277777777777775mm....or 1mm = 2.8346456692913384 pt (1/72 inch point system)
245381
5.9736×1024 kg\n\ncheck the link for more info
402707
The site has too much information to put on here but check at http://www.equipped.com/earthqk.htm
451278
here is a calculator for determining planet and orbit sizes to scale.  i hope this helps.\n\nhttp://www.exploratorium.edu/ronh/solar_system/
18543
The relative size of Earth, Jupiter, Sun = 1 : 10 : 100
1098405
just ask  what board size do u need but u don't want to look stupid so try 7.75by 31.75
43171
are you serious? i hope this is being asked by a child (and a small child at that!!) and not an adult.!!!\n if you grew at the speed at light then 0.13 seconds after conception you'd be the size of the earth!
426077
The circumference of the earth at the equator is 24,901.55 miles (40,075.16 kilometers).
1058940
at the very least seven inches.
613765
See in\nhttp://www.seds.org/nineplanets/nineplanets/nineplanets.html\nGood luck!
931645
first answer was close to right.\n\nthere is one qualification that seems need.\n(do this all in meters for easyness of calculation)\n\nR = radius of earth = 6400 km = 6.4 * 10^6 meters\nd = distance to horizon in meters\nh = height in meters\n\nSimilar triangles:\n\nR / d  = d / 2h\n\nd^2 = 2Rh\n\nd = sqrt (2Rh)\n\n....  works pretty well for small "h"\n\neg 2 meters height =>\n\nd = sqrt (2 * 6.4 * 10^6 * 2 ) = sqrt (25.6 * 10^6) \n=5.06 km\n\n[[ online calculator gives 5.5 km ]\n\nmine's a LOT easier to use ...\n\nI use "2h" instead of "h" for similar triangles\nbecause of discrepancy of right angle at the\nradius from eearth-center to observer and\nr-a from earth-center to distant object.\n[intersection happens "h" below earth surface]\n\nthat sounds confusing but I can't draw a picture\n\nmy eqn is easiest possible tho ..
674138
Very similar to Earth's size.
254490
probably pretty darn big
1011745
The mass of the earth has been calculated as 6 x 10^24 kilograms. (Mass is a little different than weight, but let's not get into that.) Here's a link on how this calculation was done.
751259
Earth is roughly 8,000 miles in diameter, and 25,000 miles in circumference.
391671
Well, try to estimate ...\n\nCertainly, much-much-much-much less than even 2 to 549th power ... of course you cannot count, but you can make a very rough estimate.\n\nThe radius of Earth is roughly 6000-7000 km, so its surface area is ~ 10^8 km^2 = 10^16 cm^2\n\nNow go out in your back yard and count the average number of grass blades/cm^2 ... let's say you get 1/cm^2\n\nAs not all of the Earth is covered with grass, its unlikely that there are much more blades than 10^16. So 12900459358983340354 is definitely WRONG. But it's also unlikely that there are less than 10^13. Now I know that 10^16 is 1000 times greater than 10^13, but even with such rough estimates, we do know more than before doing these calculations, don't we?