Respuesta :
(a)
The velocity of the meteorite just before hitting the ground is:
[tex]v=20 km/s=20000 m/s[/tex]
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
[tex]\Delta K = \frac{1}{2}mv^2= \frac{1}{2}(3.4 \cdot 10^6 kg)(20000 m/s)^2=6.8 \cdot 10^{14}J [/tex]
(b) 1 megaton of tnt is equal to [tex]1 MT=4.2 \cdot 10^{15}J[/tex]
To find to how many megatons the meteorite energy loss [tex]\Delta E[/tex]
corresponds, we can set the following proportion
[tex]1 MT: 4.2 \cdot 10^{15}J=x: \Delta E[/tex]
And so we find
[tex]x= \frac{\Delta E}{4.2 \cdot 10^{15}J} = \frac{6.8 \cdot 10^{14}J }{4.2 \cdot 10^{15}J} =0.162 MT[/tex]
So, 0.162 megatons.
(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of [tex]\Delta E=0.162 MT=162 kT[/tex]. So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
[tex]1:13 kT=x:162 kT[/tex]
And so we find
[tex]x= \frac{162 kT}{13 kT}=12.46 [/tex]
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.
The velocity of the meteorite just before hitting the ground is:
[tex]v=20 km/s=20000 m/s[/tex]
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
[tex]\Delta K = \frac{1}{2}mv^2= \frac{1}{2}(3.4 \cdot 10^6 kg)(20000 m/s)^2=6.8 \cdot 10^{14}J [/tex]
(b) 1 megaton of tnt is equal to [tex]1 MT=4.2 \cdot 10^{15}J[/tex]
To find to how many megatons the meteorite energy loss [tex]\Delta E[/tex]
corresponds, we can set the following proportion
[tex]1 MT: 4.2 \cdot 10^{15}J=x: \Delta E[/tex]
And so we find
[tex]x= \frac{\Delta E}{4.2 \cdot 10^{15}J} = \frac{6.8 \cdot 10^{14}J }{4.2 \cdot 10^{15}J} =0.162 MT[/tex]
So, 0.162 megatons.
(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of [tex]\Delta E=0.162 MT=162 kT[/tex]. So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
[tex]1:13 kT=x:162 kT[/tex]
And so we find
[tex]x= \frac{162 kT}{13 kT}=12.46 [/tex]
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.