Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Chemistry4me on 31/05/2009 04:15:59

It's a bit quite at the moment so how about having a physics problem of the week? [:X]
Here's an easy one for you:
A man spins a 0.5 kg stone on the end of a 2 meter string which has a breaking strain of 25 N. He whirls the stone 2.45 m above the ground. Eventually the string breaks and the stone flies off horizontally. Ignoring gravity [:I] air resistance, how far does the stone land from his feet?

Ignoring gravity? I guess it would hit a building or a tree or something, don't think it would have enough energy to leave the Earths atmosphere before getting slowed down and stopped by air resistance. So, it wouldn't land at all?

OPPPS [:I], I meant air resistance! [:D]

May we assume the Earth is flat so that the path taken is parabolic not elliptical ?

Yes. Parabolic.

Ignoring the correction that should be made for the fact that the string is not horizontal when it breaks I calculate that the distance will be 22.66 meters

That's not what I got [:\]. Maybe the question will be better illustrated with a diagram.
[diagram=461_0]
The string is parallel to the ground when it breaks.

My fault I worked with a one meter string, the way you posed the question was fine its just that I can't read!
PS I assume it is the stone that is 2.45m high when the string breaks not the hands of the operator.

PS I assume it is the stone that is 2.45m high when the string breaks not the hands of the operator.
I am not sure what you mean.

The string will never be horizontal there will always be a drop from the operators hands to the height of the stone.
This will have a small effect on the breaking strain which I have decided can be ignored but the height of the stone when it is released is highly relevent.
I await some other answers after my wrong result, this is a very commen cause of failure in exams not reading the question

Is there no one out there who understands simple physics and can do arithmetic ?.
I will post no further calculations until I see what other solutions are offered.

Looks like everyone is on holiday. [:)]

100m?

What? 100m! No. A hammer throw doesn't even go that far. [:)]

ok, yes I am wrong. can u tell me how to calculate the speed of the object just before the string breaks?

F_{c} = mv^{2}
r

thanks.. :) what about 70.7107m

Err...that's not what I got. How are you calculating this?

i got the final velocity by:
√(F_{c}*r) = v
m
giving me 10m/s
then i just calculated the time which the object takes to touch the floor by:
Δy=1/2(V_{i}+V_{f})t
Which gave me: .707107s
at the end I just multiply those values... [:[]

by the way, if u Karen see this post. please fix my smiley.

She can't. That embarrassed smiley is permanently broken.

can she modify it ????
here try with this one:
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fimg2.photographersdirect.com%2Fimg%2F12366%2Fwm%2Fpd901510.jpg&hash=a0a64d4285dba64c811229a353f95cbc)

Your calculations are correct except for the last one, 0.707 x 10 doesn't equal 70.7 m. [:)] Then there is another step... [:)]

lol,, (https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fimg2.photographersdirect.com%2Fimg%2F12366%2Fwm%2Fpd901510.jpg&hash=a0a64d4285dba64c811229a353f95cbc)

what is the next step?

+2

No. Pythagoras.

I dont get it. I think that I need a lecture about rotating objects,, no two just in case.

√(7.07^{2} + 2^{2}) = 7.347 m

:) sweet. thanks now I feel less stupid
[diagram=463_0]
[:D] now I better go to sleep some [:P]

Buenas noches

D=GT^2/2........
Time for stone to fall 2.45 meters = ((2.45*2)/9.8)^0.5 = 0.7071 secs
F=MV^2........
Velocity of stone when string breaks = ((2*25)/.5)^0.5 = 10 M/sec
Therefore distance travelled before grounding = 7.07 meters.
calculation completed 17.55 31/05/09
My first attempt at a calculation not being able to remember the basic physics I used Kepler instead of Newton and somehow a factor Pi crept in.

Then again I failed to read the question and omitted the Pythagoras bit!

So you think this thread is worth continuing, or should it die? [:D]
If in the affirmative, I shall post another one next weekend.

It is surprising that out of the four answers submitted by well educated corresponds none were correct despite having access to Google and Wiki.
Most did not read the question and some like me did not remember the basic physics or made arithmetical errors.
I wonder how a group of Chinease or Japanese students would have done.
Please find some more problems like these to keep us on our toes and stop us worrying about black holes and dark matter etc.

stop us worrying about black holes and dark matter etc.
There's nothing better than getting back to the basics! [:)]

made arithmetical errors.
[::)] [:P] [;D]

Lol  I get 4.994 metres

If we want greater precision allowance can be made for the reduced radius of the path that the stone takes, this is reduced to 1.961 meters by sagging and and feeds into the last part of the calculation as to how far from the operators feet the stone grounds.
This reduces the distance to 7.338 meters

:) sweet. thanks now I feel less stupid
[diagram=463_0]
[:D] now I better go to sleep some [:P]
Hold on, Pythagoras theorem can be only use in right triangles how do we now that the paths of rock are perpendicular?

Gratuitous photo to demonstrate my lightning reflexes (and stupidity)...
[ Invalid Attachment ]
I took this photo by sticking the camera lens through the safety cage,
(if the guy in the skirt had mistimed his throw I would have had to have the camera surgically removed from my skull).
BTW the weight here is 28lbs, they also fling a 56lb version.

erickejah
It is assumed that the circular path traced by the stone before its release is horizontal, if it was not that would be a whole new ball game.
In the case of the 'Highlands games competitor' the same circle would not be horizontal and the 'hammer' would be released when it was moving up.

It's looks like it's about to knock him out.
[ Invalid Attachment ]

I might be away for the next few days so I'll post another one now:
Some painters are working in a warehouse. They have a uniform plank which is resting on two supports. The plank is 4 m long. It has a mass of 22 kg. The two legs that support the plank are 0.5 m from either end.
[diagram=465_0]
Calculate the support force on the plank at A if a painter of mass 60 kg sits 0.75 m from A and another painter of mass 75 kg sits at a distance of 0.8 m from B.

No one wants a go?

I am very busy at the moment with holiday preparations but the problem seems pretty simple, just calculate the proportion of painters weight that falls on each support which will be partitioned in accordance to the ratio of their distances from the supports and the contribution of the plank itself and add them together
PS
Why did you not have some heavy equipment sitting right at the end of the plank, that would have made it more interesting ?.

The answer is:
Taking moments about B.
F_{A} x 3 = 588 x 2.25 + 735 x 0.8 + 215.6 x 1.5 = 2234.4 N
F_{A} = 744.8 N

cool, another one please

It's looks like it's about to knock him out.
That is more likely in the "weight for height" contest: (throw 56lbs weight over polevaulttype bar)...
[ Invalid Attachment ]

I thought if you threw anything horizontaly then gravity would be 9.81 m/s/s. The problem seems to boil down to the earth being bannana shaped. I don't think the mass has much to do with things.
I've been in a few highland games and the people don't come across as people that do these kinda sums. They are more worried about avoiding the highland dancers and getting the noise of the bagpipes out of their ears.

cool, another one please
Alright, see if you can work this one out then.
Phugoid oscillations.
A boy is flying his radiocontrolled aeroplane (mass = 3.67 kg) at a speed of 36.0 ms^{–1} in a straight and level flight at an altitude of 50.0 m. The plane suddenly experiences a region of turbulence, causing it to lift several metres above its original altitude. The amplitude of the vertical oscillations is 4.56 m. Calculate the maximum upward force acting on the plane during the time that the plane is oscillating.

F=ma
F=(3.67Kg)(9.8m/s^2)
F=35.966N
PE1=mgh
PE1=(3.67Kg)(9.8m/s^2)(50m) PE2=(3.67Kg)(9.8m/s^2)(54.67m)
PE1=1798.3J PE2=1962.3J
[:P]
F = Fx
PE1 PE2
35.966N = Fx
 
1798.3J 1962.3J
Fx=39.246
ΔF= FxF → 39.24635.966 = 3.28N
This is my assumption without researching about Phugoid oscillations. I will try to look that theory ASAP to do it right. I wonder how different that answer will be [???] [:D] [:D]

I did not get 3.28 N.

Can you show me the path, dont I need the AoA, or the time in which it changes?

Have you done your research?

kind of,,

It look's something like this:
[ Invalid Attachment ]
But the plane experiences an upward force of 3.67 x 9.8 N just to keep it from falling, therefore F_{max(total)} = 38.45 N

nice :)

another one?

John is sitting in a trolley and Sam pulls him along with a rope. John's mass is 65 kg, the trolley's mass is 11 kg. The tension force in the rope attached to the trolley is 95 N, and the rope is at an angle of 45^{o} to the ground. There is a 35 N friction force on the trolley. Calculate the size of the trolley's acceleration.

0.883883 m/s^2

Err...don't think so.

o yeah there is some friction [:\]
Fx=95cos45
Fx=67.1751N
35N of friction:
67.1751N35N= 32.1751
F=ma
32.1751N=(76Kg)a
a= 0.423357m/s^2

Right you are.

what is the next one?

Haven't found it yet! [:D]

Calculate the heat energy that is produced by the 3.40 Ω resistor in one minute when the diode is connected in the circuit.
[ Invalid Attachment ]

Since the diode is not being polarized and it will never be. the circuit becomes to be a series circuit with a total resistance of:
5.2Ω+3.4Ω = 7.9Ω
the current going trough the circuit is I=12V/7.9Ω → I = 1.51899A
the power in the resistor can be calculated by P=(I^2)*R → 1.51899A*3.4Ω = 5.16456W
then the total power dissipated in 1 minute is equal to:
Energy(watt hours)= Power (Watts)x Time (Hours)
W=(5.16456W)x(1/60)
W=86.076mWh
I think [:P] [:)]

the power in the resistor can be calculated by P=(I^{2})*R → 1.51899A*3.4Ω = 5.16456W

again... !!!!!!!!!!!!!!!!!!!!!!!!! this makes me mad.. let me fix it
___________________
ok its 130.749 mWh
[:D]

What's with all these mistakes? [:)]

i dont know i guess that i have a split brain.. lol

it may be the flow affected by the used of too much Facebook...
anyhow, is this answer close to the one that u had?

ok its 130.749 mWh
Um...how did you get that?

Can I get an answer in Joules?

would it be 7.84492J ?

P = E/t

so,
7.84492=E/60
E = 470.695J

Yes. That was what I got.

thanks for baring with me. :)

when are u going to put the next one, I will double check my answers this time.. :)

Are these questions too easy for you?

they are okay I guess, I only need resolve them earlier in the day. because fatigue affects my reasoning sometimes...

Here:
Tom was looking at the image of newsprint using a concave lens by holding it close to the page of the newspaper. The actual height of the words on the newspaper is 3.0 mm. The image produced is 1.0 mm high when the lens is held 3.0 cm from the print on the newspaper. Calculate the focal length of the lens.

Here:
Tom was looking at the image of newsprint using a concave lens by holding it close to the page of the newspaper. The actual height of the words on the newspaper is 3.0 mm. The image produced is 1.0 mm high when the lens is held 3.0 cm from the print on the newspaper. Calculate the focal length of the lens.
15 mm.
Enlargement: x/y = p/q > x/y = 3/1 (enlargement is negative if object and image are on the same side).
3 = p/q > q = p/3 = 10 mm (negative because the image is on the same side of the object).
1/p + 1/q = 1/f > 1/30  1/10 = 1/f > f = 15mm (negative because is a diverging lens).
http://en.wikipedia.org/wiki/Lens_(optics)
Edit: for a better explanation of the terms, x is the height of the object, y of the image; p is the object's distance from the lens' centre and q the image's distance from the lens' centre (negative if on the same side of the object).

Yes, correct, you're on to it lightarrow. [:)]

nice, I would have done the same exactly thing,,, [:D] [:D] JK lightarrow is the man [:)]

hit me up with other one

hit me up with other one
From a height R falls (from zero speed) a mass point which follows a curved trajectory of 1/4 circumference (so the total lenght is π/2*R).
From the same height R falls, at the same instant of time, from zero speed, a mass point along a stright inclined plane which is long π/2*R as well.
Which mass point arrives first?

For point masses they both arrive at the same time, for real life rolling balls or cylinders the whole situation is much more complex as some of the potential energy is stored in the rotation of the objects.

For point masses they both arrive at the same time,
Why?

I don't know about the specific two shapes in the original problem but,although you can say that they will arrive with the same speed, whatever the profile of the track (initial PE will always convert to the same KE), the time taken for the journey is not independent of the profile. Just imagine a track that starts off almost horizontal and then plunges down. It could take an hour to get to the edge of the first section.
From that, I conclude that it is unlikely that the two paths would take the same time. I'd have to do the integral for the quadrant journey  someone else can do it  or have they?
I imagine the fastest journey time would be for a vertical drop, followed by a very small radius curve and then a horizontal section to the end.

I'd guess that the curved track mass will hit first. My reasoning is that the curved track has most of its acceleration occurring at the beginning. It starts from freefall, and ends up rolling horizontally. The straight track has constant acceleration all the way down. Since the speed it has at any point depends on the integrated acceleration it has seen in the past, putting high acceleration at the beginning would give it a higher speed early on, which would make it cover the distance faster.

For point masses they both arrive at the same time,
Why?
I would like to say that the acceleration in both balls is the same in the Y direction, they both end the journey at the same time. this is my assumption in a frictionfree world and a flywheel effect free environment.
s

I would like in view of what Sopiecentaur said to withdraw my answer (that both would arrive at the same time) I now see it to be incorrect, it seems that despite the generous suppositions made a more detailed analysis is required.
A straight vertical drop would of course take the shortest time while on the inclined plane the effective value of G would be divided by the tangent of the angle of inclination relative to the vertical i.e if it was horizontal G would be zero.
I assume for the curved track we have to calculate the effective value of G for each infinitesimal point on the track, a job for those well versed in calculus.

For point masses they both arrive at the same time,
Why?
I would like to say that the acceleration in both balls is the same in the Y direction, they both end the journey at the same time. this is my assumption in a frictionfree world and a flywheel effect free environment.
s
No, the y component of acceleration is NOT the same.
It is g/√2 on the straight, diagonal path but it starts as g and ends up as zero on the circular path.
jp's qualitative analysis has to be correct. It needs to be done with yer actual Maths if you really want the right answer but my two 'extreme' examples show the way things must be going.
Just read your post, syphrum. Thanks I am just doing the sums  so far it seems that there is a factor of √2 !!! involved.
Edit typo

In many ways I see a similarity to this problem with that of calculating the time it takes for the gravity train to reach the centre of the earth (which also defeated me!)

The gravity train is 'just' simple harmonic motion  the attractive force is proportional to the distance from the centre and the 'constants involved just involve the density of the particular planet. You can either accept that the sums work or slog through it. What fascinated me about that problem was that it reveals that a piece of dust would have the same time period moving through a hole in a large granite rock as long as there were nothing else around and the rock was not tumbling (of course)
Anyway, I have had an interesting hour or so  which I should have spent doing something useful  working out some answers.
For the diagonal path.
The acceleration, a, is g/√2 and the distance to travel is l√2.
The equation of motion gives s = at^{2}/2
So l√2 = gt^{2} /(2√2)
Giving
t = 2√(l/g)
For the circular path it is easiest to consider the angle from the centre of the circle – starting at θ = 0 and going up to 90^{o}
At any point on the path, the tangential speed, v, is l dθ/dt and the 'imposed' tangential acceleration is
g cos(θ) so you can write what is happening to the mass at that point
The acceleration will be dv/dt,
or ld_{2}θ/dt^{2}
and you can write the equation as
g cos(θ) = ld_{2}θ/dt^{2}
I was messing around for ages before I realised that this is the diff equation for a simple PENDULUM –durrrr.
The period of a pendulum with small displacements (swings) is given by 2π√(l/g) and we are dealing with a quarter of a period so the time would be
t = (π/2)√(l/g) for a small displacement. (Edit: that's a pi  not an n  it isn't clear on my browser)
Or 1.57√(l/g)
BUT, for large displacements, the sums are much harder and I had to look it up. You can’t work it out simply and you have to get into “elliptic functions of the first kind – aaaargh! Wiki tells us that, for a 90^{o} swing, there is an 18% increase in time, giving
t = 1.85√(l/g)
Please check for accuracy but I think it's ok and shows that the straight path takes longer.
I then looked at what happens if the mass falls vertically, then, with the acquired speed (after a tight curve) it travels the horizontal distance.
The vertical drop takes time t_{v} = √2 √(l/g) (same as the circular path, apparently  and a bit counter intuitive) and its final speed is √2gl, so it takes an extra time
t_{h} = l/√2gl or √(l/2g)
Total time
t = t_{v} + t_{h}
t = √(2l/g) + √(l/2g)
or √(l/g) (√2 + √(1/2))
so
t = 2.12√(l/g)
Which, is greater than for the circular path.
Somewhere in there there must be an optimum.
There must be some Maths wizards out there who could find it?

The vertical drop takes time
...
t = 2.12√(l/g)
Which, is greater than for the circular path.
Of course you considered the fact that the total path in this case is greater than (π/2)R, isnt'it?

Sophiecentaur and jpetruccelli have given the right answer: the point mass in the circular path arrives first.
The following is my solution.
Said:
m = mass of the pointmass
s = curved coordinate from the initial point
v = velocity = ds/dt
R = maximum height of the point mass = radius of the circumference of the curved path
h(s) = height of the mass point with respect to the lowest point of the path
E = total energy = mgR
V = gravitational potential energy = mgh(s)
Kinetic energy = E  V = mgR  mgh(s)
we have:
(1/2)mv^{2} = E  V > v = ds/dt = Sqrt[2(E  V)/m]
> dt = ds*Sqrt[m/2(E  V)] = ds*Sqrt[1/2g(R  h(s))] = ds/Sqrt[2g(R  h(s))]
Let's compare h(s) in the two cases:
Circular path: h(s) = R[1  sin(s/R)] computed writing s = R*α, where α = angle between the instantaneous vector (from the circumference' centre and the point mass) and the horizontal
Straight path: h(s) = R  (2/π)s (similar triangles)
You can see that h(s) for the circular path is always smaller than h(s) for the straight path:
R[1  sin(s/R)] < R  (2/π)s > sin(s/R) > (2/π)s/R
that is: sinx > (2/π)x 0 < x < π/2
This last inequality is true; it's simple to see it drawing a graph of the functions sinx and (2/π)x.
So h(s) is lower, for the circular path, then R  h(s) is greater and so dt = ds/Sqrt[2g(R  h(s))] is lower.
Integrating dt, it follows that the total time is lower for the circular path.

That's a novel way of looking at it.
But does it show that the third option I suggested takes longer? It would suggest that, as the height is lost quicker, it would get there faster so the inequality wouldn't always always apply.
So what is the shape of fastest path from top to bottom?

I take it you mean the shape of a path of length h*2^.5 a solution although not very elegant mathematically would be an animation program where the shape of the curve could be pushed about similar to gamma control animation.
When I was at Siemens Hell thirty years ago we had a Basic program where we had to land a Moon landing vehicle with limited fuel with blasts of power from its engines I see certain similarities, the solution was to expend half the fuel in the first burn, half of the remainder in the second and so on.
congratulations on your solution.

I calculate that if the object takes an "L" shaped path to traverse the same distance it takes 85.4% of the time the object on the inclined plane takes but to reach the same horizontal position takes 105.9% of the time.
The path length of the circular route is only 78.54 (pi/4) the length of the "L" shaped route that opens up the possibility that it could be quicker although it is 111.1% the length of the inclined plane route but the initial acceleration will be greater.

That's a novel way of looking at it.
But does it show that the third option I suggested takes longer? It would suggest that, as the height is lost quicker, it would get there faster so the inequality wouldn't always always apply.
So what is the shape of fastest path from top to bottom?
I had posted a response, but then realized my math was off, so here goes again.
The constraints on the problem are that total length traveled = R*Pi/2, and that the mass drop a distance of R total. I think that your numbers are off in the last case you posted. Here's what I get:
Step 1: The mass falls vertically the distance R. The time for this is
tv=Sqrt[2]*Sqrt[R/g],
and the final velocity is
v=tv*g=Sqrt[2gR]
Step 2: The mass maintains its velocity and finishes the total distance horizontally. The remaining distance is R*Pi/2R=R*(Pi2)/2. The time to travel this distance is just
th=R*(Pi2)/(2*tv)=R*(Pi2)/(2*Sqrt[2*g*R])=(Pi2)/[2*Sqrt(2)]*Sqrt[R/g]
The total time is then
t=tv+th={(Pi2)/[2*Sqrt(2)]+Sqrt(2)}*Sqrt[R/g]~1.82*Sqrt[R/g].
This is the fastest of the three times, as it should be.

But What is the problem, does it have to travel the same distance in its chosen path or does it have to reach the same position horizontaly as the body moving down the incined plane ?.

I have just stumbled upon the circuit diagram problem, was it deliberately intended that the diode was polarized in the the non conductive direction or was this an error?.
Was the purpose of the question just to test ones understanding of circuit drawing conventions as the arithmetic involved was negligible ?.

Jp
I don't see how pi can come into an expression for a vertically falling body.

From a height R falls (from zero speed) a mass point which follows a curved trajectory of 1/4 circumference (so the total lenght is π/2*R).
From the same height R falls, at the same instant of time, from zero speed, a mass point along a stright inclined plane which is long π/2*R as well.
I read the original problem to mean that all tracks were required to be R*Pi/2 in length, hence the Pi entering.

I thought that the starts and destinations were the same.
No wonder we disagree!!
Ah well.

Is there any program like Multisim for physics?

I had posted a response, but then realized my math was off, so here goes again.
The constraints on the problem are that total length traveled = R*Pi/2, and that the mass drop a distance of R total. I think that your numbers are off in the last case you posted. Here's what I get:
Step 1: The mass falls vertically the distance R. The time for this is
tv=Sqrt[2]*Sqrt[R/g],
and the final velocity is
v=tv*g=Sqrt[2gR]
Step 2: The mass maintains its velocity and finishes the total distance horizontally. The remaining distance is R*Pi/2R=R*(Pi2)/2. The time to travel this distance is just
th=R*(Pi2)/(2*tv)=R*(Pi2)/(2*Sqrt[2*g*R])=(Pi2)/[2*Sqrt(2)]*Sqrt[R/g]
The total time is then
t=tv+th={(Pi2)/[2*Sqrt(2)]+Sqrt(2)}*Sqrt[R/g]~1.82*Sqrt[R/g].
This is the fastest of the three times, as it should be.
Yes, this is the fastest path, you are correct.

That's a novel way of looking at it.
But does it show that the third option I suggested takes longer? It would suggest that, as the height is lost quicker, it would get there faster so the inequality wouldn't always always apply.
So what is the shape of fastest path from top to bottom?
It is just the one you guessed, your third option; the inequality holds for that path, furthermore, h(s) is minimum in that case.

You are right, of course and it solves the problem, as presented.
I read it wrong but the fastest journey from A to B (i.e. displacement) is a more likely scenario to deal with than how quickly you can travel a certain distance.
I guess, if you had to drop a ball down a fixed length of tubing from a given height above ground, the original question would deal with that scenario.

You are right, of course and it solves the problem, as presented.
I read it wrong but the fastest journey from A to B (i.e. displacement) is a more likely scenario to deal with than how quickly you can travel a certain distance.
I guess, if you had to drop a ball down a fixed length of tubing from a given height above ground, the original question would deal with that scenario.
Yes. Note (but probably you already know it) that if, instead, you have to find the quicker path between *two fixed points* A and B with the second at lower height than the first, it would be a different problem, known as the 'Brachistochrone' problem:
http://en.wikipedia.org/wiki/Brachistochrone_curve
which solution is an hyperbolic cosine a cycloid.
To solve this last problem we can use variation calculus and the resulting Lagrange equation. Only one function y(x) solve the equation (the cycloid).
I say this because I tried to use the same method for the problem I posted and I found that there is no solution to the resulting equation, that is no function h(s) for which the Functional F(h): h(s)>T (T = total time) is stationary.
For a normal function f(x), if it's never stationary, you can say that its maximum and minimum values (if they exist) must be at the border of the domain. But which is the border of the domain for the functional F(h), if we can use the same reasoning (not sure of it)?
Intuitively it seems that the function h(s) for which the functional is minimum is that one which graph is horizontal for a lenght (π/2  1)R and then vertical for a lenght R (kind of opposite of the quicker path). Does it mean that those two functions h(s) represent the 'border' of the domain of F(h)?

You've gone and spoiled my evening now. I shall have to start thinking!

An LCR series circuit is connected to an AC voltage source:
[ Invalid Attachment ]
The voltages across all the components were measured and this data was obtained:
[ Invalid Attachment ]
After looking at the data it was argued that the voltmeter was faulty because the inductor voltage was larger than the source voltage. Is the voltmeter faulty? Why? Why not?

Insufficient data is given to calculate frequency etc but this is quite normal behavior for a AC circuit incorporating capacitors and inductors, at the resonant frequency the impedance of the capacitor and inductor will cancel each other out and the current will be determined by the resistor while the voltage across the reactive components will be determined by their reactance times the current.
It is assumed that the source impedence of the AC supply is zero.

Does no one want to calculate the phase angle or the "Q" ?.

Does no one want to calculate the phase angle or the "Q" ?.
I have computed the quantities ωL/R = 75/36, ω^{2}LC = 31/15, ωRC = (4/45)√481, but I made some error in the computations because the numbers are not consistent.
Edit: Found and fixed the error. The correct values are:
ωL/R = 25/12
ω^{2}LC = 25/9
ωRC = 4/3
> Q = (1/R)√(L/C) = 5/4 (Exact values).

Using phasor notation:
The triangle of voltages has 7.2 real. 9.6, resultant of Xl  Xc. Square and add gives you (12)squared, which you'd expect.
The angle between input voltage and voltage across resistor is 53degees.
Say there's1A flowing, to make it easier.
I reckon that the resonant frequency will be 0.6 of the test frequency (that would be when 15ω = 5.4/ω, when Xl = Xc).
So we would know the impedance at resonance (7.2) and Modulus(impedance) at the test frequency (7.2+j9.6) and also the ratio of frequencies is 1:1.67.
That should allow you to work out the Q but I need to think about how.
Anyone else, please?
[Edit for clarity of maths]

Given that the resonant frequency f1 equals the original freqency f0 times 0.6 (Sophiecentaur post 5643) the impedance of the inductor is now 0.6 of its previous value.
At f0 the ratio of resistance to impedance was 15 to 7.2 now it is 1.25 which is the "Q" value at this frequency.
Can we go further do we have sufficient data to work out the value of the resistor ?
Arithmetical error corrected!

I think my head just exploded after all of that.

As we've assumed the current yo be 1A, the value of R is 7.2 ohms.

Didn't we already know that R was 7.2 ohms?

The exact value of Q = (1/R)√(L/C) is 5/4 (see post up).

How can we include √(L/C) in our calculations when we are told nothing about them ?.
We are told nothing about the value of R we can only assume a convinient value.
In my opinion the only information we can derive from the data given is the Q factor (1.25) and the phase angle of the current (53.13°).

How can we include √(L/C) in our calculations when we are told nothing about them ?.
Not understood this one.
We are told nothing about the value of R we can only assume a convinient value.
In my opinion the only information we can derive from the data given is the Q factor (1.25) and the phase angle of the current (53.13°).
Exactly. We have 3 equations and 4 unknown. Solved them we have the parameters I wrote, from which we can only find what you said.

Why does E_{k} = 0.5mv^{2} but E = mc^{2}
How are the two equations related? What happened to the 'm'?

E=mc^2 is derived from geometric considerations and is relevent as velocities approach c E=0.5*mv^2 belongs to the world before Einstein and Lorentz and does not take into account the changes that take place in M as the velocity increases.

Why does E_{k} = 0.5mv^{2} but E = mc^{2}
They are both false [:)]
How are the two equations related? What happened to the 'm'?
E_{k} = 0.5mv^{2} is valid only for bodies with non zero mass and at non relativistic speeds; E = mc^{2} is valid only at zero speed.
The correct one, valid for all speeds and with massive or nonmassive bodies (in special relativity) is:
E^{2} = (cp)^{2} + (mc^{2})^{2}
where p = momentum. If m = 0 (photons, gluons, ecc.) then E = cp. At zero velocity p = 0 so E = mc^{2}.
Kinetic energy is total energy minus rest energy:
E_{k} = E  mc^{2}

How can we include √(L/C) in our calculations when we are told nothing about them ?.
Not understood this one.
We are told nothing about the value of R we can only assume a convinient value.
In my opinion the only information we can derive from the data given is the Q factor (1.25) and the phase angle of the current (53.13°).
Exactly. We have 3 equations and 4 unknown. Solved them we have the parameters I wrote, from which we can only find what you said.
Yes  the circuit can be scaled up or down in Impedance, according to taste and availability of components. It's just the ratios that count.

E_{k} = 0.5mv^{2} is valid only for bodies with non zero mass and at non relativistic speeds; E = mc^{2} is valid only at zero speed.
The correct one, valid for all speeds and with massive or nonmassive bodies (in special relativity) is:
E^{2} = (cp)^{2} + (mc^{2})^{2}
where p = momentum. If m = 0 (photons, gluons, ecc.) then E = cp. At zero velocity p = 0 so E = mc^{2}.
Kinetic energy is total energy minus rest energy:
E_{k} = E  mc^{2}
Ooh [:o] (showing my ignorance here) That's a new one! I must really remember that.

This question has me scratching my head, I can't seem to get the 'correct' answer. [:I] [:\] Maybe I'm not understanding something. It goes something like this: A uniform beam 2.0 metres long is known to weigh between 400 N and 500 N. Using a spring scale, S and a pivot block, P, an experiment was made with the beam balanced horizontally on P and S to enable the weight to be calculated. The spring scale reads 85 N when P is 0.75 m from one end, what is the weight of the beam?