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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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If both waves have the same wavelength, then the amplitude of
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3 years ago
Heptane and water do not mix, and heptane has a lower density (0.684 g/mL) than water (1.00 g/mL). A graduated cylinder contains
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Given that the density of heptane is

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7 0
1 year ago
A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr
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From this, taking R=0.08205746\frac{atm.l}{mol.K},  we have:

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T_{1}=\frac{P_{1}V_{1}}{nR}

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3 years ago
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