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goldenfox [79]
2 years ago
5

A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle

is 3.6 m long.
Physics
2 answers:
Ksju [112]2 years ago
8 0

Thhhhhhhhhhhhhhhhhhhhhhhhheeeeeeeeeeeeeeee answer is 1.56

Nastasia [14]2 years ago
5 0

Answer:

F=35651.04N

Explanation:

Let's start writing all the data from the question.

We have the mass ''m'' of the cannonball :

m=7.5kg

We have the final speed ''Vf'' of the cannonball :

Vf=185\frac{m}{s}

And we have '' Δx '' which is the total distance that the cannonball travelled inside the cannon muzzle :

Δx = 3.6m

If we want to find the average net force applied to the ball, we need to find the acceleration of the cannonball (because we already have the mass of the cannonball).

We can use the following equation in order to find the acceleration :

Vf^{2}=Vi^{2}+2.a.Δx (I)

Where ''Vi'' is the initial speed.

Where ''a'' is the acceleration

If we replace all the data in the equation (I) (And assuming Vi=0 because that is the initial speed of the cannonball before the cannon is shot) :

(185\frac{m}{s})^{2}=0^{2}+2.a.(3.6m)

a=\frac{34225\frac{m^{2}}{s^{2}}}{(2).(3.6m)}=4753.47\frac{m}{s^{2}}

a=4753.47\frac{m}{s^{2}}

Now, we can use the Newton's second law of motion to find the force ''F''

F=m.a

F=(7.5kg).(4753.47\frac{m}{s^{2}})

F=35651.04N

Where ''N'' is the unit Newton a force unit.

We find that the average net force applied to the ball is F=35651.04N

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Given

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Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat (U_{g}), in joules, is equal to the sum of the final translational kinetic energy (K), in joules, and work losses due to air resistance (W_{l}), in joules:

U_{g} = K +W_{l} (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l} (2)

Where:

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g - Gravitational acceleration, in meters per square second.

h - Initial height of the cat, in meters.

v - Final speed of the cat, in meters per second.

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v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }

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The speed of the cat when it hits the ground is approximately 7.586 meters per second.

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Answer:

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\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j}  )

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