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SurpriseDog
SurpriseDog
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Let's say the bullet is a standard .308 Winchester round with a diameter of 7.85mm which means the bullet is encountering:

$\pi * 3.925^2 * 1000 = 48,400$ cubic mm of air.

The density of air at 20 degrees C is $1.2 KG / m^3$ which means the bullet is encountering around $0.058$ grams of air per meter of its flight path.

Via $E=MC^2$, a total of 0.058116 grams of matter is equal to 522000000000000010440000000000 Joules of Energy which is more than the 3000 or so Joules that was imparted to the bullet on launch. At this rate the bullet won't even make it 1 meter (or 1mm) away from the rifle before the explosion on its tip bring it to a sudden halt and then annihilating itself a terrific mushroom cloud. Maybe next time, we should save the antimatter weapons for Space!

Let's say the bullet is a standard .308 Winchester round with a diameter of 7.85mm which means the bullet is encountering:

$\pi * 3.925^2 * 1000 = 48,400$ cubic mm of air.

The density of air at 20 degrees C is $1.2 KG / m^3$ which means the bullet is encountering around $0.058$ grams of air per meter of its flight path.

Via $E=MC^2$, 0.058 grams of matter is equal to 5220000000000000 Joules of Energy which is more than the 3000 or so Joules that was imparted to the bullet on launch. At this rate the bullet won't even make it 1 meter (or 1mm) away from the rifle before the explosion on its tip bring it to a sudden halt and then annihilating itself a terrific mushroom cloud. Maybe next time, we should save the antimatter weapons for Space!

Let's say the bullet is a standard .308 Winchester round with a diameter of 7.85mm which means the bullet is encountering:

$\pi * 3.925^2 * 1000 = 48,400$ cubic mm of air.

The density of air at 20 degrees C is $1.2 KG / m^3$ which means the bullet is encountering around $0.058$ grams of air per meter of its flight path.

Via $E=MC^2$, a total of 0.116 grams of matter is equal to 10440000000000 Joules of Energy which is more than the 3000 or so Joules that was imparted to the bullet on launch. At this rate the bullet won't even make it 1 meter (or 1mm) away from the rifle before the explosion on its tip bring it to a sudden halt and then annihilating itself a terrific mushroom cloud. Maybe next time, we should save the antimatter weapons for Space!

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SurpriseDog
SurpriseDog
  • 4.2k
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  • 23
  • 36

Let's say the bullet is a standard .308 Winchester round with a diameter of 7.85mm which means the bullet is encountering pi * 3.925**2 * 1000 = 48,400:

$\pi * 3.925^2 * 1000 = 48,400$ cubic mm of air.

The density of air at 20 degrees C is 1.2KG / m^3$1.2 KG / m^3$ which means the bullet is encountering around 0.058$0.058$ grams of air inper meter of its flight path.

Via E=MC^2$E=MC^2$, 0.058 grams of matter is equal to 52200000000000005220000000000000 Joules of Energy which is more than the 30003000 or so Joules that was imparted to the bullet on launch. At this rate the bullet won't even make it 1 meter (or 1mm) away from the rifle before the explosion on its tip bring it to a sudden halt and then annihilating initself a terrific mushroom cloud. Maybe next time, we should save the antimatter weapons for Space!

Let's say the bullet is a standard .308 Winchester round with a diameter of 7.85mm which means the bullet is encountering pi * 3.925**2 * 1000 = 48,400 cubic mm of air.

The density of air at 20 degrees C is 1.2KG / m^3 which means the bullet is encountering around 0.058 grams of air in its flight path.

Via E=MC^2, 0.058 grams is equal to 5220000000000000 Joules of Energy which is more than the 3000 or so Joules that was imparted to the bullet on launch. At this rate the bullet won't even make it 1 meter (or 1mm) away from the rifle before annihilating in a terrific mushroom cloud. Maybe next time, we should save the antimatter weapons for Space!

Let's say the bullet is a standard .308 Winchester round with a diameter of 7.85mm which means the bullet is encountering:

$\pi * 3.925^2 * 1000 = 48,400$ cubic mm of air.

The density of air at 20 degrees C is $1.2 KG / m^3$ which means the bullet is encountering around $0.058$ grams of air per meter of its flight path.

Via $E=MC^2$, 0.058 grams of matter is equal to 5220000000000000 Joules of Energy which is more than the 3000 or so Joules that was imparted to the bullet on launch. At this rate the bullet won't even make it 1 meter (or 1mm) away from the rifle before the explosion on its tip bring it to a sudden halt and then annihilating itself a terrific mushroom cloud. Maybe next time, we should save the antimatter weapons for Space!

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SurpriseDog
SurpriseDog
  • 4.2k
  • 4
  • 23
  • 36

Let's say the bullet is a standard .308 Winchester round with a diameter of 7.85mm which means the bullet is encountering pi * 3.925**2 * 1000 = 48,400 cubic mm of air.

The density of air at 20 degrees C is 1.2KG / m^3 which means the bullet is encountering around 0.058 grams of air in its flight path.

Via E=MC^2, 0.058 grams is equal to 5220000000000000 Joules of Energy which is more than the 3000 or so Joules that was imparted to the bullet on launch. At this rate the bullet won't even make it 1 meter (or 1mm) away from the rifle before annihilating in a terrific mushroom cloud. Maybe next time, we should save the antimatter weapons for Space!