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A 1040 kg artillery shell is fired from a battleship. While it is in the barrel of the gun, it experiences an acceleration of 2.2 × 104 m/s2.m=1040 kga = 2.2 × 104 m/s2Part (a)  What net force is exerted on the artillery shell before it leaves the barrel of the gun (in Newtons)? Part (b)  What is the magnitude of the force exerted on the ship by the artillery shell in Newtons?

Question

A 1040 kg artillery shell is fired from a battleship. While it is in the barrel of the gun, it experiences an acceleration of 2.2 × 104 m/s2.

m=1040 kg

a = 2.2 × 104 m/s2

Part (a)  What net force is exerted on the artillery shell before it leaves the barrel of the gun (in Newtons)?

Part (b)  What is the magnitude of the force exerted on the ship by the artillery shell in Newtons?

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Step 1

Given:

m = 1040 kg

a = 2.2×104 m/s2

Step 2

(a) Calculating the net force is exerted on the artillery shell before it leaves the barrel of the gun:

Mass of the shell, m 1040 kg
Acceleration of the shell, a 2.2x104 m/s2
According to Newton's second law, F = ma
Hence, force exerted on the shell
F = (1040kg)x(2.2x10*m /s')
F=2.3x10N
She
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Image Transcriptionclose

Mass of the shell, m 1040 kg Acceleration of the shell, a 2.2x104 m/s2 According to Newton's second law, F = ma Hence, force exerted on the shell F = (1040kg)x(2.2x10*m /s') F=2.3x10N She

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Step 3

(b) Calculating the magnitude of the force exerted...

According to Newton's third law of motion, Every action has an equal an opposite
reaction. Hence, while leaving the shell will exert an equal but opposite force on the
ship
Therefore, Magnitude of force on the ship, F2.3x 10 N
shp
help_outline

Image Transcriptionclose

According to Newton's third law of motion, Every action has an equal an opposite reaction. Hence, while leaving the shell will exert an equal but opposite force on the ship Therefore, Magnitude of force on the ship, F2.3x 10 N shp

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Newtons Laws of Motion

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