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snow_tiger [21]

3 years ago

10

A 1600kg cannon fires a 5kg cannonball horizontally. The exit velocity of the cannonball is 80m/s and the barrel length is 2m. W

hat is the average acceleration of the cannonball before it leaves the barrel?

1 answer:

evablogger [386]3 years ago

7 0

Answer:

a = 1600 m / s²

Explanation:

For this exercise we use the kinematics relations,

v² = v₀² + 2 a x

where v₀ is the initial velocity of the bullet, which as part of rest is zero, for the distance (x) we can assume that the gases accelerate along the entire trajectory of the cannon x = 2m

a = $\frac{v^2}{2x}$

let's calculate

a = $\frac{80^2}{2 \ 2}$

a = 1600 m / s²

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Given data,

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A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling

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From the question,

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Use newton's law to explain the vertical acceleration of a projectile

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Answer:

Explained below

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3 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

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The length of the hannging mass is $l_h = \frac{3}{4} l$

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The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

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$F_x =T cos (60)$

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$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

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3 years ago