• 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)

A2 Physics | Post your doubts here

Messages
155
Reaction score
65
Points
38
A geostationary satellite moves above the equator in the same direction as earth's rotation with a period of 24 hours. It orbits 36000 km above the surface of the Earth.
Thanks i got it :p
 
Messages
23
Reaction score
3
Points
13
E = -dV/dx - use the gradient of electric potential graph to plot the graph for electric field strength.
okay..and on the decreasing part of the graph(V against x) ..would the gradient would be a curve or a line with positive gradient??
 
Messages
155
Reaction score
65
Points
38
A sinusodial carrier wave has amplitude 12 volts and frequency 600 kHz.
The frequency of the carrier wave changes by 25 kHz per volt.
The carrier wave is to be used for the transmission of a signal of frequency 3 kHz and amplitude 2 volts.

For the frequency modulated carrier wave, state

1) the amplitude
2) the maximum frequency
3) the minimum frequency
4) the number of times per second that the frequency of the carrier wave changes from the maximum to minimum and then back to maximum value.


how to solve part 4 ?
 
Messages
772
Reaction score
149
Points
38
A sinusodial carrier wave has amplitude 12 volts and frequency 600 kHz.
The frequency of the carrier wave changes by 25 kHz per volt.
The carrier wave is to be used for the transmission of a signal of frequency 3 kHz and amplitude 2 volts.

For the frequency modulated carrier wave, state

1) the amplitude
2) the maximum frequency
3) the minimum frequency
4) the number of times per second that the frequency of the carrier wave changes from the maximum to minimum and then back to maximum value.

how to solve part 4 ?
It changes 3000 times per second because the frequency of signal is 3kHz. The frequency of carrier wave changes in synchrony with displacement of the signal. . .
 
Messages
2,619
Reaction score
293
Points
93
WELL for the last row.....if ice is less dense than water this means that in water molecule wud be closer compared to the that in ice......and so P.E has decreased and thus U shud decrease.....bt ms says it increased.........and also ice has larger volume than that water according to info provided so work shud be done on the system.......
MS says U as +ve ...Q as +ve and W as 0
 

Attachments

  • delta U.PNG
    delta U.PNG
    73.6 KB · Views: 4

omg

Messages
626
Reaction score
3,628
Points
253
im nt abl to post the link PLEASE smbdy help in q.3 (b)(ii) w08 p4!!! plsss
 
Messages
23
Reaction score
3
Points
13
On Fig. 4.1, mark a position of the pivot P for the piston to have
(i) maximum speed (mark this position S),
(ii) maximum acceleration (mark this position A).
N/B : Please give the explanation for each case !!! :)
qaws.jpg
 
Messages
803
Reaction score
1,287
Points
153
im nt abl to post the link PLEASE smbdy help in q.3 (b)(ii) w08 p4!!! plsss

wel i can guess some ;)

it has been said in the above part of the question that the needle is oscilating with a maximaum(total) ditance of 22mm.

and we know that the formula fr measuring the displacement(x) of the oscillating object is X= Xo(coswt)

here Xo is the maximum displacement frm mean position!!

so in the case y =acoswt
a is the maximum displacement (amplitude)

total distance as stated is 22mm so the amplitude will be itx half (22/2)
=11mm :)


fr second part we are to calculate the angular velociTy, w
the frmula is w = 2πf

w= 2π(4.5) f= 4.5 as stated in question w= 28.3 Ans
 

omg

Messages
626
Reaction score
3,628
Points
253
w07 q1 part b (ii) and s08 q3 part (c)(ii) PLEASE HELP :( (idk sth is wrong wid my pdf reader im nt able to open the ppr dats y couldnt giv the links)
 

omg

Messages
626
Reaction score
3,628
Points
253
wel i can guess some ;)

it has been said in the above part of the question that the needle is oscilating with a maximaum(total) ditance of 22mm.

and we know that the formula fr measuring the displacement(x) of the oscillating object is X= Xo(coswt)

here Xo is the maximum displacement frm mean position!!

so in the case y =acoswt
a is the maximum displacement (amplitude)

total distance as stated is 22mm so the amplitude will be itx half (22/2)
=11mm :)


fr second part we are to calculate the angular velociTy, w
the frmula is w = 2πf

w= 2π(4.5) f= 4.5 as stated in question w= 28.3 Ans

its givn y=acos wt in the qs when actually it is y=asin wt?????????????? isnt it??
 

omg

Messages
626
Reaction score
3,628
Points
253
ANd pleaseeeeeeeeeeeeeeeeeeeeeeeee smbody explain me about the signal-to-noise ratio!! please please pleaseeeeeeeeeeeeeeeeeeeeee
 
Messages
2,619
Reaction score
293
Points
93
signal to noise ratio represent the number of times the signal power must be larger than noise power in order to distinguish between signal or noise ....if this ratio becomes tooo low signal is lost into the noise
 
  • Like
Reactions: omg
Messages
803
Reaction score
1,287
Points
153
its givn y=acos wt in the qs when actually it is y=asin wt?????????????? isnt it??


well for finding displacement from the mean position in a simple harmonic motion.

we use the formula X= X0sinwt (this is used when no displacement took place at time, t = o ) means x=o at t=o.
where As
we use the formula X = Xocoswt (this is used when no displacement took place at time, t = o ) means x=Xo at t=o.

so both formulas are correct fr calculating displacement, the thing is that they r 2 b Used at diff. occasions.:)

sinwt will be used when the graph is starting frm origin( o, 0)
coswt will be used when the graph is starting frm maximum displacement ( 0, Xo) hope u gt it??
 
Top