Question #210455

An asteroid with a diameter of 11km and a mass of 4.3 x 1015 kg impacts the earth at a speed of 31414m/s, landing in the Pacific Ocean. If 1.23% of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of 12°C), what volume of ocean water in L) will be boiled away by the collision? Assume ocean water is just plain water.


Expert's answer

0.5kmv2=ρRV0.5(0.0123)(4.3⋅1015)(31414)2=(103)(2.26⋅106)VV=1.15⋅1010 L0.5kmv^2=\rho RV\\ 0.5(0.0123)(4.3\cdot10^{15})(31414)^2\\=(10^3)(2.26\cdot10^{6})V\\V=1.15\cdot10^{10}\ L


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