if a bomb radius is , say, one kilometre, does that mean the circular area affected would actually be two kilometres across?
211870
It is how far the bomb blasts.\n\nAn example of usage:\n\n"Any enemy within the bomb's radius at the time of blast will be killed"
172855
yes you're right. force1 x distance1 = force2 x distance2 \nif we have a fulcrum
339527
area of circle is A = pie r^2.\ndifferentiating with respect to t, we get,\ndA/dt = pie * 2*r*( dr/dt)\nnow, we know, the circumference is 2*pie*r. let us take 2*pie*r as C.therefore,\n2*pie*r  = C   and \nrate of change of radius dr/dt is - 0.1 cm/sec....(note the minus as its decreasing)\ntherefore rate of change of area in terms of  circumference is\ndA/dt = - C * (0.1 cm/sec)\ntherefore, answer b is correct.
418840
No...\n\nAll calculations are based off of an impact explosion model developed for determining the effect of Bunker Buster artillery.\n\nThat type of impact would be no diferent than what was seen when battleships lobbed volkswagen sized shells into the earth...  Only the size and velocity of the projectile are increased to reflect the meteor...
383391
Depends on the thickness.  A two-foot radius tree is most likely impenetrable and a bullet won't kill it.  It will however, lodge itself in the tree.  A 6-inch radius tree is most likely not bulletproof and the bullet will go through it.  Lastly, as far as I know, no tree can deflect a bullet, if that's what you were wondering.
269939
Distribution of force over area.
91985
The diameter is 3476 km so it's radius is half that.  That's 1.738 x 10^6 meters.
272427
You probably need more information...there can be multiple cylinders inscribed in a sphere with a radius of 5.
316512
False: If the radius is doubled, then its circomference is multiplied by 4.  (The circumference is just doubled.)\nThe circumference is\nC = 2pi*r\nr = 5, so \nC = 2pi*5 \nC = 10pi = 31.4159\n\nIf C = 44, then\n44 = 2pi*r\n44/(2pi) = r\nr = 7.0028
69462
110,860 sq kilometers (that's 42,803 square miles)\n\nStraight from the CIA Factbook (link below)
103896
I believe the bounds are where the numbers are deviated but still close. Say, in the problem, 11 cm is measured to the nearest centimeter. So, it may not be 11cm. It could be 10.6 cm, 11.3 cm, or even 10.51294803 cm. But they'll all be 11cm in the measurement. \n\nSo for the question, if it's asking for the minimum possible area:\nLet l be length and w be breadth. \n10.5 <= l < 11.5\n8.5 <= w < 9.5\nThe lower bound would be composed of the shortest length and breadth. \nA = 10.5(8.5) = 89.25 cm^2
391437
If its distance from the center is less than the radius, it's inside.\n\nYou can find the distance from the center using your handy pythagorean theorem.
475412
Yes,, actually we have a formula to build a Bomb that is a 500 megaton cobalt salted Nuclear weapon that if detonated could crack the core of this planet like a China plate in a shooting gallery, rip away half this plantes atmosphere, and knock this planet out of orbit around the sun,,pretty bad stuff huh?
406300
In theory, it will never stop, so I'll assume that when it bounces less than 1 mm it is finished.  Total distance travelled will be 26.99221155 meters.  Start with the 3 m drop and add the 2.4 m bounce and drop and continue the sequence until the bounce is less than 1 mm.  Here's my spread sheet:\n\n3000----------3000\n2400----------7800\n1920---------11640\n153614712\n1228.817169.6\n983.0419135.68\n786.43220708.544\n629.145621966.8352\n503.3164822973.46816\n402.65318423778.77453\n322.122547224423.01962\n257.698037824938.4157\n206.158430225350.73256\n164.926744225680.58605\n131.941395325944.46884\n105.553116326155.57507\n84.4424930126324.46006\n67.5539944126459.56804\n54.0431955326567.65444\n43.2345564226654.12355\n34.5876451426723.29884\n27.6701161126778.63907\n22.1360928926822.91126\n17.7088743126858.32901\n14.1670994526886.6632\n11.3336795626909.33056\n9.06694364726927.46445\n7.25355491826941.97156\n5.80284393426953.57725\n4.64227514726962.8618\n3.71382011826970.28944\n2.97105609426976.23155\n2.37684487526980.98524\n1.9014759 ------------26984.78819\n1.52118072 -----------26987.83055\n1.216944576----------26990.26444\n0.973555661----------26992.21155\nOk, Excel columns don't work well when copied here, so I've added spaces to a few entries at the top and bottom.\n\nGuess spaces didn't work either.  Maybe the dashes will stay.
505579
I can't find anywhere that says exactly on this.  However I could find that it tops out at about 600m above the surrounding planes.  Also it is aprox 3000m in diameter.  So if it is a perfect cone, which it is not, it would be 14,681,680 square meters surface area.  Not sure how much this helps you.
162176
It would depend on the map.  There should be a 'scale' on it someplace (usually in a corner) that will tell you.
315622
Not considering the spheric harmonic of high orders I would have to tell you that my physics spirit tell the surface area  is what counts and there you could have k0.2Pi.R=2Pi <=>k0=1/R\nk1=1/(2R), k2=1/(distance of the 3 points more away from each other)........................