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Yes. read your text book, student :) -J.J.
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He got fired.
Both objects are acted on by the force of the expanding gases in the bore. The forces forward and backward have to be equal. The motion they produce ... the forward motion of the shot and the rearward motion of the cannon ... are in inverse proportion to their masses, so that the linear momentum after the shot is the same as the linear momentum before the shot, namely zero.
no
Yes. read your text book, student :) -J.J.
"For every action, there is an equal and oposite reaction." The cannonball is pushed out of the barrel at high speed. This pushes the cannon in the opposite direction. That is recoil. The heavier the cannonball, and the faster it is pushed, the more the cannon recoils.
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Yes, the forces are balanced if the explosives in the cannon, the cannon and the cannon ball are all included in the system. The Echem is transferred to Eint in the form of heat from friction as the cannonball rubs the inside of the cannon barrel, and Ev in the form of vector motion of both the cannonball moving forward and the cannon kicking backwards. Therefor, no energy enters or leaves the system
A cannon typically makes a loud booming sound when it is fired. This sound is due to the release of compressed gas propelling the cannonball out of the barrel at high speed.
The moon has no atmosphere and has less gravity than the earth. That means that a cannonball fired on the moon will travel further.
He got fired.
That depends on basically three things: the angle at which the cannon is fired, the velocity of the projectile, and the acceleration of gravity. The maximum range of the cannon is achieved with a firing angle of 45 degrees to the horizontal. But the maximum height is achieved when the firing angle is 90 degrees to the horizontal, that is, when the cannon is pointing straight up into the air. (Which can be very dangerous to the artillery personnel firing the guns!) So, if we assume that the cannon is on Earth, not some other planet, and fired straight up into the air, can we determine how high the cannonball will go? Yes, if we know the velocity of the cannonball as it leaves the end of the cannon's tube. Once the cannonball leaves the end of the tube, it begins to slow down because of the acceleration of gravity. We can use the energy equations to calculate the maximum height of the cannonball. We know that kinetic energy is defined by the equation Ek = mv2/2. We also know that potential energy (due to altitude) is defined by the equation Ep = mgh. Equating the two, we get mv2/2 = mgh. Rearranging the terms to solve for h, we get: h = v2/2g. (Note that g = 9.8 m/s2 = 32.2 ft/s2.) So, let's say the cannon has a muzzle velocity of 1000 meters per second. The cannonball, therefore, has an initial velocity of 1000 m/s before it starts to slow down. Plugging 1000 into the equation above and solving for h, we calculate the theoretical maximum height as 51,020 meters. In practice, however, the cannonball will not achieve anywhere near that height because of air resistance, which has a tremendous effect at such high speeds.
he was hired and fired on the same day
Both objects are acted on by the force of the expanding gases in the bore. The forces forward and backward have to be equal. The motion they produce ... the forward motion of the shot and the rearward motion of the cannon ... are in inverse proportion to their masses, so that the linear momentum after the shot is the same as the linear momentum before the shot, namely zero.
throught the use of pressure/fire/gun powder
A cannonball fired horizontally and one dropped from the height of the muzzle simultaneous with the shot will hit the ground at the same instant, provided only that the ground under the muzzle and the ground where the shot lands are at the same elevation, i.e. the shot was not fired off the edge of a cliff or into the side of a mountain. To solve this kind of problems, it often helps to separate the movement, or the speed, into vertical and horizontal components. In this case, the vertical component of the speed is the same.