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In Nuclear War How Long Does It Take For Radiation And Gama Rays To Travel 60 Miles

How does nuclear radiation after an atomic blast work?

I always believe that after a nuclear explosion that there is this nuclear fallout, but isn't there a way to get rid of it? or is it going to be around for a long time. I remember hearing a weapons expert on talk radio say that U.S. Nuclear weapons now are more advanced to the point where during a nuclear explosion, all the radioactive materials burn up, thus almost reducing the Nuclear Fallout to none. Plus, I remember him saying that after a Nuclear blast, the way to get rid of the radiation is to get out a Geiger Counter with some protective gear and pick up all the radioactive materials left over. I mean how is Nagasaki and Hiroshima habitable now?

What speed does nuclear fallout travel after blast?

How long would it take fallout to travel 100 miles? Please don't give the simple answer that it just depends on the wind speed I doubt fallout would only travel 20mph if the wind speed was 20mph. I want to try to escape across the border to canada have about 20 hours of driving before I get to my safe zone Good God bless!!

How long does light take to reach the Earth from the Sun?

Let us try to analyze this mathematically. The earth moves around the sun in an elliptical orbit. The length of the semi-major axis is 1.496 x 10^8 km while it’s eccentricity is 0.0167.Since the apogee is the point where the earth is farthest from the sun, time taken by light to reach the earth when it is at the apogee will be maximum while the exact opposite will be valid when the earth is at the perigee.Thus distance of apogee=Semi-major axis x (1 + eccentricity)=1.496 x 10^8 x (1 +0.0167) =1.52098 x 10^8 kmThe distance of perigee=Semi-major axis x (1 - eccentricity)=1.496 x 10^8 x (1 -0.0167) =1.471016 x 10^8 kmSpeed of light is 3 x 10^5 km.Thus time taken by light when the Earth is at apogee =(Distance from apogee)/3 x 10^5=507 s=8 min 27 sec.Time taken by light when the Earth is at perigee =(Distance from perigee)/3 x 10^5=490 s=8 min 10 sec.At any other point on the orbit, the time required varies from 8 min 10 sec to 8 min 27 sec

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