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Andrei [34K]
3 years ago
13

Sir Marvin decided to improve the destructive power of his cannon by increasing the size of his cannonballs. Sir Seymour kept hi

s cannonballs the same size, but improved his powder to provide more velocity. a) Which knight will have the more destructive cannon? Why?
Physics
1 answer:
maria [59]3 years ago
5 0
We really can't tell from the given information. 
We don't know HOW MUCH Marv enlarged his cannonballs,
or HOW MUCH faster Seymour's balls became.

If we assume that they both, let's say, DOUBLED something,
then Seymour accomplished more, and the destructive capability
of his balls has increased more. 

I say that because the destructive capability of a cannonball is
pretty much just its kinetic energy when it arrives and hits the target.
Now, we all know the equation for kinetic energy.

                K.E.  =  (1/2) (mass) (speed-SQUARED) .

We can see right away that if Marv started shooting balls with
double the mass but at the same speed, then they have double
the kinetic energy of the old ones.

But if Seymour started shooting the same balls with double the SPEED,
then they have (2-SQUARED) as much kinetic energy as they used to.

That's 4 times as much destructive capability as before.  

So we can say that when it comes to cannons and their balls and
smashing things to bits and terrorizing your opponents, if making
a bigger mess is better, then more mass is better, but more speed
is better-squared.
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Part a)

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Part b)

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Part c)

v_y = \sqrt{Rg/3}

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X = \frac{13R}{4}

Explanation:

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initial speed = v_i

angle = \theta_i degree

Now the two components of the velocity

v_x = v_i cos\theta_i

similarly we have

v_y = v_i sin\theta_i

Part a)

Now we know that horizontal range is given as

R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}

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Part b)

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Part c)

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Part d)

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Part g)

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so maximum range is

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