18 Nov 2008 10:02:09 UTC

Topic 194041

(moderation:

I really want to study an MSc in Medical Physics next year and have found a couple of courses I want to apply for. One of them is in BC Canada and they want me to sit some sort of equivalency test in physics as part of my application.

It's been 12 years since I last studied physics.

For the last 5 years my level of physics has been at demonstrating basic phenomena using, ramps, trolleys, mild electrocution, elastic bands, sticks, springs, old lab equipment that often won't work, standing waves by microwaving cheese etc..so my fundamentals are good (pun intended) but I have forgotten the vast majority of what i learned at university.

I downloaded the trial paper and most of it was "oh..I used to know that..now i haven't a clue".

Physics Test Preparation Materials - pdf doc

This is not good.

Basically I don't think I can swot it up in time to apply for the BC course, but if I work I may be able to remember enough to cope with the UK course and not get hopelessly lost.

I'm going to attempt to work through this paper (mainly for my own revision purposes), but I need help. I know that the Einstein boards are full of really brainy and kind hearted physicists who'd love to explain things to me. :))

So the first 10 questions:

Question 1 was easy,

I need to check my working on Q2 because i came up with the wrong answer..but i think i know what i did wrong.

3: easy

4: easy

5:easy

6:ok

7: Any links to help me revise this can someone help? I can't find my old text book on it :(

7: Easy

8,9 and 10. I am rusty on. Is there anyone there who is good on electric fields who can help me with this one?

I'll get to the other 90 questions once I'm happy with the first 10. This could take a while...they get much harder.

Physics is for gurls!

Language

Copyright © 2024 Einstein@Home. All rights reserved.

## Calling all Physicists - help with physics needed!

)

Also, if there are any medical physicists here who can suggest specific areas I should focus on, I'd be grateful.

Physics is for gurls!

## RE: Question 1 was easy,

)

Oh was it :-)

Actually I'd bet that 95 % of the population would pick a wrong one, I guess answer (B) would be the most popular (confusing velocity and acceleration).

CU

Bikeman

## I wish you the best of luck,

)

I wish you the best of luck, I think you're gonna need it....

Waiting for Godot & salvation :-)

Why do doctors have to practice?

You'd think they'd have got it right by now

## RE: RE: Question 1 was

)

Well I've just been teaching Simple Harmonic Motion to my year 13s..so it would be a bit bad if I got that question wrong :D

Physics is for gurls!

## RE: I wish you the best of

)

:(

Physics is for gurls!

## So..any help with

)

So..any help with these?

If it is an ideal gas does that mean it's reversible?

Physics is for gurls!

## #1. E - Pendulum swings in a

)

#1. E - Pendulum swings in a circular arc, thus acceleration ( though provided by combination of gravity and line tension ) is directed towards centre at all points.

#2. 33.3 revs per minute = 33.3/60 revs per second = (33.3/60) * 2 * PI = radians per second = w = 3.4871649

if ds infinitesimal arc length travelled in dt , and dQ is infinitesimal angle change in dt then if R is the radius

ds = R * dQ and ds/dt = R * dQ/dt = R * w

v = tangential velocity = ds/dt = wR

a = centripetal acceleration = (v^2)/R = (w^2) * R

this is provided ( in the maximal case of verge of slipping ) by friction

F = m ( 0.3 )

( as the normal to the plane is equal to the weight and the co-efficient of static friction gives the frictional component parallel to the plane which is proportional to that normal )

thus

m * 0.3 = m * (w^2) * R

ie. R = 0.3/(3.4871649) ^ 2) = 0.024 ( about )

#3. D - see Kepler

#4. If v0 is velocity of 2m prior to collision and v1 is velocity of 3m after, then conservation of momentum

2 * m * v0 = 3 * m * v1 -> v1/v0 = 2/3

KE( before ) = 2m * v0^2 / 2

KE( after ) = 3m * v1^2 / 2

KE( after )/KE( before ) = (3/2) * (2/3)^2 = 2/3

fraction lost is = 1 - 2/3 = 1/3 ( answer C )

#5. D - equipartition theorem and you get a kT per degree of freeedom.

#6. B - If it's isothermal there's no work done by the gas - it had to be provided from elsewhere to expand it without a temperature drop - the average kinetic energy per particle is constant - and ideal and monatomic implies no 'within particle' or 'between particle' potential energy changes. If it's adiabatic there's no energy to contribute to expansion other than from the gas, so that work will be non-zero. Quasi-static implies reversible....

#7. B - option A is for North/South, heaven knows where the others are from.

#8. D - Opposites attract, so A, B & C are out. Infinite and grounded means charge from elsewhere can come in but from a long way away, conducting means the field lines are perpendicular to the plane. The nett effect is as if there is a charge of -Q an equal distance on the other side of the plane with the plane removed. Mirror image.

#9. A - Symmetry, all the vectors will sum to null. Could be gravity with masses - Kepler Rosette.

#10. Energy = E = q * V ( q is charge, V is Voltage )

capacitance = C = q / V ( by definition )

E = C * V * V

Now capacitors in series combine ( like resistors in parallel )

1/Ct = 1/C1 + 1/C2 + 1/C3 ..... + 1/Cn

Ct is total of combo, C1 ..... Cn the separate capacitance values.

For 3 and 6 microFarads 1/Ct = 1000000 * ( 1/3 + 1/6 ) = 1000000 * ( 1/2 )

ie. Ct = 2 microFarads

so E = 2 * 300 * 300 / 1000000 = 18 / 100 = .18 J ( option B )

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

## Ok, here's the next ten .....

)

Ok, here's the next ten ..... :-)

#11. D and E are out as the lenses are convex ( if concave one can have a virtual image on the same side as the object ). So how far to the right of the

rightmost lens?

If you have a focal length of F, with D1 and D2 being the distances from ( midline within ) the lens then

1/F = 1/D1 + 1/D2

if either D1 ( or D2 ) go to infinity then 1/F = 1/D2 ( 1/D1 ), thus the image is at D2 = F ( or D1 = F ). This defines the meaning of focal length ie. at

what point do infinitely distant rays ( parallel to symmetry axis ) converge?

So for the left lens with object at 40cm .....

1/20 = 1/40 + 1/D2

-> 1/D2 = 1/40 and D2 = 40 which is 10cm to the right of the second lens. You still use the above formula again but with care in interpretation.

1/10 = 1/D1 + 1/10

1/D1 = 2/10

D1 = 5cm

now this is on the right side of the second lens also. Basically you have rays from the first lens already converging before they hit the second's

surface. If from infinity ie. parallel they would have converged at 10cm. So the extra convergence has brought the image in closer than 10cm.

#12. If the object was at the focal point ( F/II ) then the rays would go to infinity on the left side ( definition of focal length ). Now the object is

closer in to the mirror than F and rays will make a greater angle to the normal when hitting the mirror's surface. This will give diverging rays to

the left which converge to a virtual image on the right. Virtual means the light never actually comes/goes there but the rays behave as if they did!

Thus V is the only possible point - option E.

#13. I think it was Airy or Rayleigh who did the diffraction analysis and came up with

sin(angular separation) = wavelength / aperture

- plus or minus a bit ( factor of 1.2-ish? ) as one has to 'define' when a diffraction minimum from one source overlaps/obscures an adjacent maximum from the other. For small angles the sin(angle) = angle. So

aperture = wavelength / angle = 600 * 10^(-9) / 3 * 10^(-5) = 200 * 10^(-4) = 2 * 10^(-2) = 2 cm

So I'd take option B.

#14. With the source at the face the detector is capturing all radiation emitted to that side, a hemispheres worth. The 8cm depth/thickness is sufficient to grab all that is incident when source is close, assume the same with source at distance. So at 1m away a surface of 8cm radius intercepts what fraction of a hemisphere? [ Area of a sphere = 4 * PI * R^2 ]

Total area of hemisphere = (4 * PI * 1^2)/2

Total area of surface = PI * (0.08)^2

Total area of surface/Total area of hemisphere = ((0.08)^2)/2 = 8 * 10^(-4)/2 = 4 * 10^(-4)

Thus option C

#15. A - although not as accurate as the others ( ie. reflecting the true value ), they have the least spread of results.

#16. To use a sample's statistics to guess the error in one's estimate of the population mean you use:

(variance of sample)/(size of sample)

meaning that you'd like the sample mean to estimate the ( unknown ) population mean to within some degree of confidence. The central limit theorem says that

pretty well every type of population distribution ( various assumptions! ) will behave like this.

The mean is 2 and the variance is 8 for this sample. We want

8/N 100 * 8 = 800

So that would be option C.

#17. They all identify 15 electron spots, so that's good. The right one is option B. [ The orbital filling order doesn't get squirrelly until the 4th row of the periodic table. ]

#18. A singly ionised Helium is a hydrogen-like where the energy goes like 1/Z^2 ( Z is nuclear charge ). For hydrogen ( Z = 1 ) E = 13.6 eV hence for Z = 2, E = 13.6 * 4 = 54.4. Option D. The double ionization value is a red herring ( except that the singly ionised value ought be less than that, but all of A thru E are ).

#19. Assuming we are talking basic Hydrogen to Helium fusion, then 4 separate protons wind up being bounds as 2 protons + 2 neutrons in Helium. The electrons

are irrelevant, so is talk of atoms vs. ions. So that's option B, the others don't balance nucleons except for option D but that's not the primary source in

our Sun.

#20. E - 'braking radiation' as the electrons get deep into the well near the metal nuclei and flick off photons going around the sharp turn. Unrelated to

electron shells except that you need energy to penetrate them to get to the nucleus. Thus smooth and continuous as all sorts of offsets from 'dead on' the

nucleus occurs.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

## Wow, the curve is getting

)

Wow, the curve is getting steeper! :-)

#21. For hydrogen the energy levels go like 1/n^2 where n is the principle quantum number. The lines are energy transitions between the levels.

Energy ratio = Lyman/Balmer = (1/1^2 - 1/2^2)/(1/2^2 - 1/3^2) = (3/4)/(5/36) = 27/5

Now energy goes like frequency which goes like inverse wavelength, thus

Wavelength ratio = 5/27 ( option B )

#22. A and B. The min and max separations will give the semi-major axis ( option E, just add them ) which will give the period ( option D, via Kepler ), and

with the maximal orbital speed ( at perigee ) will give the minimum speed at apogee ( option C, due to conservation of angular momemtum, or equal areas in

equal times if you like - same thing ). You can get the combined mass but not either separately from this study.

#23. 90 degress always by definition of circular motion. Option E.

#24. Horizontal component of velocity is constant. The vertical component starts positive and diminishes at a constant rate to zero at the top and then falls

back down ie. negative. Option C only correctly identifies both behaviours.

#25. The intro helpfully gave the moment about centre of mass for the disc of mass M and radius R as (M * R^2)/2. So one of those only for the central penny. The six surrounding ones will have that same moment but displaced to around the central axis of the middle one. The parallel axis theorem gives a simple addition here :

I(displaced) = I(centre) + M * r^2 [ here r is the displacement from centre of mass, which happens to equal 2 penny radii with this geometry ]

I(displaced) = (M * R^2)/2 + M * (2R)^2 = (M * R^2) * [ 1/2 + 4 ] = (M * R^2) * 9/2

You got six pennies and the central one

I(total) = I(central) + 6 * I(outer) = (M * R^2)[ 1/2 + 6 * 9/2 ] = (M * R^2)[ 1/2 + 6 * 9/2 ] = (M * R^2)[ 55/2 ]

Option E

#26. Get back on that one.

#27. Ditto!

#28. The dot product must be zero to be orthogonal ( forget the context - it's a vector space ).

5 * 1 + (-3) * (-5) + 2 * x = 0

5 + 15 + 2 * x = 0

2*x = -20

x = -10

Option E

Orthonormal means mutually orthogonal plus each is of length unity.

#29. Expectation/probability comes from the dot product of the wave function with itself. Orthonormal implies the cross terms are zero. It's a square root of the sum of squares.

SQRT(1/6 + 1/2 + 1/3) = 1

Option C

#30. II doesn't diminish to zero at infinity. I and III do but A/r doesn't normalise ( area under curve is infinite ). That leaves I, option A.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

## Were those questions in

)

Were those questions in English?

It's been a few years since uni physics and I was glad to get out of there. Crap, I barely made it through the electricity stuff. Those diagrams made my head hurt.

You're a better woman than I, Es.

Kathryn :o)

Einstein@Home Moderator