Physics questions HELP!

Topps

Need help with these please I have no idea where to start and they are due tomorrow, could someone help me out. Much appreciated

3. A freshly prepared sample of gold with an atomic mass number of 198 has an activity of 2.50 x 108 Bq. The half life of this isotope is 64.8 hours. It decays by beta - emission.
Write an equation for this nuclear reaction.
How many gold atoms are in the sample initially?
What is the decay constant?
How long before the activity has reduced to 15% of the initial activity?

4. The proper length of starship Enterprise is four times that of the starship Physics. Both starships travel in the same direction. Radar observations measure that both ships have the same length. If the slower ship has a speed of 0.4c, where c is the speed of light in vacuum, determine the speed of the faster ship.

iLaura

I'm assuming this is for Joe Meehan? :P

For question 4, if you go to tutorials we did a question almost exactly the same on page 50, its q4. If you use the formula
L=Lp*sqrt(1-v2/c2)

For q3 I only have the first part done so far. 198/79 Au --> 0/-1 e- + 198/80 Hg

I'm still doing the rest. Struggling majorly with q1 though

citrus burst

I'm not quite sure what you mean by equation. A mathematical equation or a chemical one. A chemical one is simple enough. Beta emission is essentially the decay of a neutron. A neutron decays into a proton with the emission of an electron and a neutrino, so from that you should be able to work out a chemical equation. Look at the periodic table.

For the next part you need to find the decay constant [Latex]\lambda[/Latex]. There is a relation between it and the half life. From there you should be able to work out the initial amount of Au with the given.

Last part you should be able to work out from what you've done.

4). The proper length is the length of an object as determined by its own reference frame. There is an equation for it, similar to Pythargoras' formula. Next work out the length contraction. You don't have enough information to work it out fully but, if you get the ratios right, you should be able to work out the velocity.

Hint 1: set c = 1
Hint 2: faster objects are contracted more.

Hope I haven't been too vague

Topps

Yes the legend himself:D

OK Question1: some of this may not be right:pac:

(I) period 60s/12 oscillations =5s

(II) Angular frequency w=2pi/T =2pi/5

(III)Max acceleration a=-w*x =(-1.26)*(200x10-3)

(IV)X =ACOS(wt) Rearrange to get t

then use V=-Awsin(Wt) filling in t

(IV) again for some reason) X=Acos(Wt) work out x

(V) X=Asin (wt) t=1/w.sin-1(x/A)

hope this helps. May be subject to many errors:D

Topps

iLuara how do you do 4 im lost with no figures

iLaura

Topps wrote: »

Yes the legend himself:D

OK Question1: some of this may not be right:pac:

(I) period 60s/12 oscillations =5s

(II) Angular frequency w=2pi/T =2pi/5

(III)Max acceleration a=-w*x =(-1.26)*(200x10-3)

(IV)X =ACOS(wt) Rearrange to get t

then use V=-Awsin(Wt) filling in t

(IV) again for some reason) X=Acos(Wt) work out x

(V) X=Asin (wt) t=1/w.sin-1(x/A)

hope this helps. May be subject to many errors:D

It's better than nothing! :P THANKS

Okay, question 3. I worked out the decay constant first as I don't see how to get the initial number otherwise.

So T(1/2) = ln2/lambda
lambda = 2.97x10^-6

The the number initially
Ro = Lambda*No
No = (2.5x10^8)/(2.97x10^-6)
No = 8.42x10^13

And I haven't the foggiest of how to work out the activity bit.

Do you still need help with q4 or are you okay with that?

iLaura

Q4 answer, hope it attached now

Topps

Sound out Luara I owe u one. Could you just explain q3 again im a bit confused

iLaura

No bothers. Mind explaining/putting up the maths for q1 IV? I haven't the foggiest what to do, even with your help :P

Topps

ya no bother thanks again

Q1 (IV) The speed when 10mm from equilibrium position, v

Ok first we have to get t and then differentiate the equation and get v

X=Acos(wt) manipulate formula to get t on its own

X/A=cos(wt)

t= 1/w.cos-1(X/A)

Okay now fill in figures

t=1/1.26.cos-1(10x10-3/200x10-3) work that out and you get

t=1.21s

Now we use
X=Acos(wt) again but differentiate it first

V(speed)=dx/dt =-Awsin(wt)

V=(-200x10-3)(1.26).sin((1.26)(1.21))

work that out and you get a negative number of -0.252ms-1

dunno about the negative

Anything else?

iLaura

I think it's only negative as it's moving away from its equilibrium point?

But nope, thats it all done now, unless you happened to miraculously get an answer for the activity in q3? :P I can finally leave the library now!

Thanks a billion for your help ^_^

Topps

If i get it ill let u know check this before the lecture tomorrow at 9 :eek:. Thanks for 3 and 4