Iron Shot
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3 pcs Tattoo Machine Guns Iron Shot Twin Coil 10 wrap $71.99 |
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2 pcs Tattoo Gun Iron Shot Twin Coil 2 Color by Express $53.99 |
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2 pcs Tattoo Gun Iron Shot Twin Coil 2 Color by Express $53.99 |
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2 pcs Tattoo Gun Iron Shot Twin Coil 2 Color by Express $53.99 |
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1 pc Red Tattoo Machine Gun Iron Shot Twin Coil Express $35.99 |
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1 pc golden Tattoo Machine Gun Iron Shot Twin Coil $35.99 |
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1 pc Black Tattoo Gun Iron Shot Twin Coil by Express $35.99 |
How long would it take 125 mm diameter iron Shot Put (weight 16 lbs.) In the fall of 417 meters in the water?
How many long would it take 125 mm diameter spherical solid cast Shot (weight 16 lbs.) to fall a distance of 417 meters in the water? The scenario I want is that it is under water and falls 417 meters of pure water. It starts at zero velocity at the exit point and Earth's gravity takes over as it falls into the water. How many seconds will have been when he crossed the milestone of 417 meters? I know the acceleration will be slower than if it were in a vacuum and the acceleration will be slowed as the shot approaches terminal velocity in water. Someone there physical know enough to figure this out. What form (s) did you use and how did you plug in the numbers? I hope I have given sufficient information to enable understanding.
Assuming that the drag on the cannonball is quite viscous drag will proportional to the speed of the cannonball in the water. The ball will then accelerate until it reaches a terminal velocity, which we call Vmax. At the point where the ball reaches terminal velocity, the upward force of drag equals the downward gravitational force of gravity. Under such conditions, the velocity obeys the equation a = v * PDO – g-When the speed reaches Vmax, the equation becomes 0 = ao *- Vmax – g then AO =- G / Vmax and if the velocity is zero at time zero, then V (t) = Vmax (e ^ (-g/Vmax * t) – 1) can be calculated how the ball will be delayed in the achieving significantly Vmax, but compared to 417 meters, the ball will fall, the delay will be negligible. The question then becomes calcuating Vmax and if the ball moves slowly enough to drag it to be viscous rather than turbulent, which is governed by another equation. This means the viscosity of water and some other things of fluid dynamics. I do not remember how to do these calculations, but I remember that the viscosity of water depends on temperature and the ball is slow somewhat as it reaches the colder, more viscous layers of water.
