• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Physics: Post your doubts here!

Messages
65
Reaction score
90
Points
28
I'm assuming that you got the answers for the previous two parts.. for part (iii),
Intensity received back= (intensity reflected at the muscle-bone boundary)e^-(23 * 4.1 * 10^-2)

[ e^-(23 * 4.1 * 10^-2) = 0.39]


intensity of the wave incident on the boundary was 0.39I, as calculated in part (ii)
0.33 of this was reflected at the boundary.
So intensity reflected at the muscle-bone boundary= .39 * .33I

answer= .39 * .33I * .39 = .050I

I hope this helps

ohh k i understand but i am confused in the end why did u multiplied .39*.33 again with .39? o_O
 
Messages
515
Reaction score
1,447
Points
153
does anyone remember what was the answer to the question about the circuit with a variable and constant resistor in parallel and the other six ohm resistor in series , the one with the 24 volt battery ? in the part where he asked to calculate the resistance of the constant resistor X ? this ques could be the difference between my getting an A or a B .. please help :( .... Tense!
p.s im talking about v 22
my ans was 3.6.. but my frnd got 6.67... now we r nt sure that which one is correct :/
 
Messages
1,983
Reaction score
3,044
Points
273
A parallel beam of light of wavelength 450 nm falls normally on a diffraction grating which has
300 lines / mm.
What is the total number of transmitted maxima?

mohammad salik
Thought blocker
Suchal Riaz
at the nth maxima angle is 90. sin90=1
so in formula we get
d=nλ ==> n=d/λ

d=1*10^-3/300=3.33X10^6
so n=3.33X10^6/450X10^-9 = 7.4
so maximum order = 7
there are 7 maxima above middle maxima and 7 below
to total 7+7+1=15
:D
 
Messages
8,477
Reaction score
34,837
Points
698
A parallel beam of light of wavelength 450 nm falls normally on a diffraction grating which has
300 lines / mm.
What is the total number of transmitted maxima?


Mohammed salik
Thought blocker
Suchal Riaz

the ans is 15
sin(90) = 1 is the largest real value that sin(theta) can have. In other words 90 degree is the limit for possible diffraction angle magnitude. Rearrange the diffraction grating formula and you have

sin(theta)=(n)lambda/d

Clearly n can only take values that result in the R.H.S. being less than or equal to one.

Note that the system is symmetric about the center line from the diffraction grating to the projected image. That means maxima will occur on both sides of the center line with the same angular spacing (and there is also a central maximum on the center line).

GOT IT ?
 
Messages
515
Reaction score
1,447
Points
153
sin(90) = 1 is the largest real value that sin(theta) can have. In other words 90 degree is the limit for possible diffraction angle magnitude. Rearrange the diffraction grating formula and you have

sin(theta)=(n)lambda/d

Clearly n can only take values that result in the R.H.S. being less than or equal to one.

Note that the system is symmetric about the center line from the diffraction grating to the projected image. That means maxima will occur on both sides of the center line with the same angular spacing (and there is also a central maximum on the center line).

GOT IT ?
i did the same n i got 7 number of maximas... while it is 15 in ms :/
 
Messages
195
Reaction score
495
Points
73
Look they said 3000 revolutions per minute
So I divided 3000 by 60*10 cuz it said 10cm wide I got 5 rounded it to 10 ._. Not sure if its correct
Look this is what i could do.
60secs-3000 revolutions
1sec-3000/60=50revolutions
T=1/f
=1/50=20ms
So if its 10ms the wave can be easily be shown on the c.r.o .LOL i asked for an answer and solved it myself :p
 
Messages
6
Reaction score
7
Points
3
coursebook chapter 19, exam style question 2d:
venus has diameter 12100 km and mass 4.87x10^24 kg
calculate energy needed to lift one kg from the surface of venus to a space station in orbit 900 km from the surface.
how to get the answer?
 
Messages
222
Reaction score
166
Points
53
Kindly make correction in the answer which I posted for your correction!
increase number of bits in digital number at each sampling ......................................M1
so that step height is reduced ................................................................................... A1 increase sampling frequency / reduce time between samples ..................................M1
so that depth / width of step is reduced ..................................................................... A1 [4] (do not allow ‘smoother output’)
I wrote it upside down,just saw it now,sorry
Thanks brother
 
Top