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An artillery shell has a mass of 75 kg. The projectile is fired from the weapon and has a velocity of 670 m/s when it leaves the barrel. The gun barrel is 2.7 m long. Assuming the force and therefore the acceleration is constant while the projectile is in the barrel, what is the force that acted on the projectile?

A) 16,650 N
B) 25,110 N
C) 55,200 N
D) 42,750 N

User Bullgare
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1 Answer

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Final answer:

The force acting on a 75 kg artillery shell fired with a velocity of 670 m/s from a 2.7 m long barrel can be found using kinematics and Newton's second law, but calculated values do not match the provided options.

Step-by-step explanation:

The question asks us to find the force that acted on a 75 kg artillery shell which was fired with a velocity of 670 m/s from a barrel that is 2.7 m long. By using the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity (0 m/s), v is the final velocity (670 m/s), and s is the distance (2.7 m), we can solve for a which is the acceleration. Once we have the acceleration, we can find the force using Newton's second law, F = ma, where m is the mass (75 kg) and a is the acceleration.

First, we calculate the acceleration: 670^2 = 0 + 2 * a * 2.7, which gives us a = 83017 m/s^2. Then, we calculate the force: F = ma = 75 kg * 83017 m/s^2, which gives the force acting on the projectile as 6226275 N. However this is not one of the provided answers, indicating a calculation error. Let's redo the calculation:

First we calculate acceleration by rearranging the equation a = v^2 / (2s). With a = (670)^2 / (2 * 2.7) we get a = 83017 / 5.4 ~ 15373 m/s^2. The force is then F = ma = 75 kg * 15373 m/s^2 = 1152975 N. This calculation still does not match any of the provided options, meaning that there might be a further error in calculation or a misunderstanding in the question. There could also be a typo in the options provided by the student.

Without additional information or clarification, it's advised to review the initial question and the steps followed to ensure the accuracy of the provided data and calculations.

User Johno Scott
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