How long will an iron cannon ball thrown from a boat take to reach the deepest point of the ocean? - if a cannon ball deepest ocean
Lowest point of the ocean is the Marianas Trench at 10,911 m (35,797 ft) below sea level. Assume that the ball is perfectly round and weighs 30 pounds = 13.6 kilograms. Making reasonable assumptions for the coefficients of friction, buoyancy, and other parameters.
To be clear .... I do not know the answer, but I wonder is :-)
Thursday, February 4, 2010
If A Cannon Ball Deepest Ocean How Long Will An Iron Cannon Ball Thrown From A Boat Take To Reach The Deepest Point Of The Ocean?
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6 comments:
It is a strange question to ask only for certain curiosity.
The mass of the projectile is 13.6 kilograms, it has a weight of about 133.4 Newtons (assuming g = 9.81 m / s ^ 2).
The strength of the peso is down ... However, because the ball is not in a vacuum and / or no point the crowd with a vibrant non-zero force acting on it is upward.
The shear force on the ball depends on the weight of liquid water (in this case), it moves.
The iron has a density of about 7874 kg / m ^ 3, so that a ball has a volume of 13.6 kg of about 1.73 E-3 m ^ 3
Under the assumption that the salt water ocean has a uniform density kg of about 1028 / m ^ 3, then the weight of water displaced is exerts a force of about 17.42 Newtons up.
Thus, the net force on (stationary) cannonball about 116 Newtons downward.
After the 2nd Newton's law
F = ma
This force will cause the ball from 13.6 kg to that at a speed of 8.53 m to accelerate / s 2 ^
Assuming that draw no "water" resistance /, then according to the formulaLA
d = 1 / 2 a ^ 2,
where the ball from rest and undergoes constant acceleration during a time t when the ball covers the distance d
The ball should reach about 50.57 seconds on the ground.
However,
Just as it would air friction when the ball was thrown into the air, it is more difficult, "water" as the ball sped along .... eventually reaches a terminal velocity.
The water will exert a force upward pull on the ball, so its acceleration is not uniform.
The resistance can be calculated as follows,
F_d C_d = * ((p * v ^ 2) / 2) * A
F_d is where the power draw is C_d the drag coefficient, p is the density of the medium that will drop the ball, v is the velocity of the ball, and A is the cross section of the ball in his direction.
In order to calculate the drag force, we need to know the drag coefficient of the ball in the water, which should be determined experimentally in general.
The ball accelerates unevenlyreaches terminal velocity in water ceases at that time, the acceleration and the ball fell into his final speed.
If more information is known (the drag coefficient), the better (I guess) / estimated duration of the ball is finally over 50.57 seconds ... probably to the tune of more than a minute that the ball does not accelerate nearly as long as in our previous mathematical model of computation.
depends on ocean currents and weather or not, landed on a whale again
2 minutes
2 minutes
With the density of iron and assuming a perfect sphere, the ball would have a diameter of about 15.4 cm. Assuming a density of sea water of 1030 kg/m3, the balance of power against gravity flotation should almost came to 113.7 N in the downward direction. I think the time of 51 seconds.
Not enough information. You need to know to send the size of the weapons themselves, their rate of fire and the type of explosive used for the ball in the path.
The slope of the plan depends on the distance.
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