Advanced Physics Help Needed [update]
-
one more..!
http://i105.photobucket.com/albums/m225/nickykeeng/album2/physics.png
ugh got another set of questions.. have to do first ~.~
-
yoz since you are in university already, you should depend on yourself. Be independent.
-
Originally posted by WoAiMeiMei:
one more..!
http://i105.photobucket.com/albums/m225/nickykeeng/album2/physics.png
ugh got another set of questions.. have to do first ~.~
hey my soln wrong ah? =(
-
a) If particle reaches x=20cm, difference in PE between equilibrium position and amplitude position = 16J
KE of particle at equilibrium position = 0.5 * 1.7 * 0.8^2 = 0.544J < 16JThus, ans is yes, will turn back before reaching x=20cm
b) U(x) = bx^2
We take the min point as U(x) = 0J
when x = 20, U(x) = 16J
so b = 0.04
At turning point KE = 0, so U(x) = 0.544J = 0.04x^2
x = sqrt (13.6) = 3.69cm -
Originally posted by eagle:
a) If particle reaches x=20cm, difference in PE between equilibrium position and amplitude position = 16J
KE of particle at equilibrium position = 0.5 * 1.7 * 0.8^2 = 0.544J < 16JThus, ans is yes, will turn back before reaching x=20cm
b) U(x) = bx^2
We take the min point as U(x) = 0J
when x = 20, U(x) = 16J
so b = 0.04
At turning point KE = 0, so U(x) = 0.544J = 0.04x^2
x = sqrt (13.6) = 3.69cmhaha.. at least i got the 0.544J part right..
-
thanks hey haha. it's like 1 peer learning experience : D
-
ok la... What course u taking? Physics or Engineering?
-
yeah doing engineering hey. sorry for the late reply.. stacked with soo much work it's not funny. /emo
got some new questions on post 1
-
Sry, I'm quite busy with my FYP thesis... it's due 4th Apr... So might not have time to help much...
-
ok ok *panic*
-
ps: is there a way to edit thread title?
-
what do u want to change it to? Moderators can do it
-
-->
"Advanced Physics Help Needed [update]"
=D
-
I see the questions I also sian... My weakest topic somemore...
I do one question first...
Chapter 33 Problem 36
Normally, the intensity is proportional too the square of the cosine of the angle between the two polariods. I remember this because of it costed me dearly when I took part in a competition in 2001.
If proportional to [cos(θ2 – θ1) ]^2, where θ1 is fixed, it means that the graph of [cos θ2 ]^2 is shifted right.
Let θ2 – θ1 = x, then (cos(x))^2 = ½ (cos(2x) + 1)
The diagram shows that it is zero when θ2 is 160 degrees. This happens when cos(2x) = -1, 2x = 180 degrees, x = 90 degrees
So θ2 – θ1 = 90 degrees when θ2 is 160 degrees
Thus, θ1 is fixed at 70 degrees.So when θ2 = 91 degrees, θ2 – θ1 = 21 degrees = x
Intensity is ½ (cos(2x) + 1) = 0.8716Answer: 87.2%
-
2001 is a long time ago o.0! i was 11 back then omg!
thankfully we're starting electromagnetism this coming week. but i have a feeling it's not going to be as easy as school...
that Chapter 33 Problem 36 is not as simple as it looks ..
i still cant get it. what would it be if it was 117 degrees?
erk this is due on tues ~,~
add: yay did a question : D
add2: and another : D
add3: as above
add4: " "
-
chapter 35, problem 13 and 21 is seriously pissing me off ):
-
chapter 35, problem 13
Original λ = 568nm
In n1, λ1 = 568/1.43 = 397.2nm
In n2, λ2 = 568/1.55 = 366.45nmIn n1, number of λ completed = 4.75μm / 397.2nm = 11.96
In n2, 3.96 μm / 366.45nm = 10.81So at exit, phase difference = 11.96 – 10.81 = 1.15 λ
Either that or answer is 0.15 λ… I’m not very which one… forgot definition liaoOk, off to gym!
-
Just discovered prob 21 is very short... so I do first:
90 degrees means λ/4 phase difference (ahead)
rA is greater by 120m, this means the phase is ‘set back’ by 0.218 λ (120/550)Hence, magnitude of phase difference at detector = 0.25 – 0.218 = 0.032 λ (with A ahead)
Hope it helps!
-
unfortunately i cant get a correct answer for those ones. all i know for prob 21 is that the units are in radians.
-
Originally posted by WoAiMeiMei:
unfortunately i cant get a correct answer for those ones. all i know for prob 21 is that the units are in radians.
1 λ = 360 degrees
0.032 λ = 11.52 degrees = 0.2 radians
-
chapt 35, Q21 is correct n ow ^^
-
questions completed/superceeded
-
got some new q's up.. will get working ^^"
-
omg chap 14 prob 22 looks easy but >__<
here's the hint
Can you find the pressure at a given depth? How is force related to pressure for a given horizontal surface area? On the vertical wall, how does pressure vary with depth along the wall? Do you see that you need to first consider a strip that has the wall's width but only a differential height dy? What is the force on the strip? Do you see that you next must integrate to find the force on the full wall (all the horizontal strips)?
-
I’m rushing a lot of work right now… I do one of the easier questions first:
Chapter 14 Prob 79
0.652mm = 0.000652m
Vol of bubble = 4/3 πr3 = 1.16*10^(-9) m3Using Archimedes principle, upward force on bubble = 998 * 1.16*10^(-9) * 9.81 = 1.14 * 10^(-5) N
So, using F = ma at that particular spot in time (vol of bubble will change as it goes higher),
M = 3.09*10^(-5) kg