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dusya [7]
3 years ago
4

The force exerted by expanding gases is what propels a shell out of a gun barrel. Suppose a force that has an average magnitude

F propels the shell so that it has a speed v at the instant it reaches the end of the barrel. Part A What magnitude force is required to double the speed of the shell in the same gun?
Physics
1 answer:
zavuch27 [327]3 years ago
6 0

Answer:

F_1 = 4F

Explanation:

We need to find a proportion between the two force. This proportion is easly identificable through the Work made for both.

Then,

W = Fl

The work done by the force on the shell = change in kinetic energy,

Fl = KE_f - KE_i

Fl = KE_f

Fl = \frac{1}{2}mv^2

We know can calculate the speed of the gun when is doubled.

F_1l=\frac{1}{2}m(2v)^2

F_1l = \frac{1}{2}m(4v^2)

F_1l=\frac{1}{2}m(4v^2)

The relation between this two states is given by

\frac{F_1l}{Fl} = \frac{\frac{1}{2}m4v^2}{\frac{1}{2}mv^2}

\frac{F_1}{F} = 4

F_1 = 4F

Therefore the magnitude force is required to double the speed of te shell in the same gun is F_1 = 4F

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Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

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Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

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Final Volume, V =4,292cm3

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