Physics: Post your doubts here!

[selrey]

Can anyone tell me when to use which equation, like what are the situations or scenarios to use each equation. this is for SHM

Considering the statement given in bracket, maybe it has to do something with whether at t=0 the displacement is max or 0. Otherwise we can use any of these. Just that if there's a part in qs which requires an eq for velocity, we will use the derivative of the eq we used for displacement. Like if sin for x then cos of v

 

[GarryTheGhost]

Can someone please explain part (c)?

lol just finished that paper do u have any idea how would i do question 6ii

 

[selrey]

lol just finished that paper do u have any idea how would i do question 6ii

From the graph R=2.5 cm. As V is constant inside a sphere

 

[anastasia grey113]

Can anyone tell me when to use which equation, like what are the situations or scenarios to use each equation. this is for SHM

if the graph of displacement against time is starting from X0 i.e. the maximum displacement, then the one with coswt should be applied otherwise, sinwt should be used.

 

[anastasia grey113]

Can someone please explain part (c)?

in c)i) draw the smoothed voltage graph as shown in our book...in it the voltage never drops to zero.
It's shape is more like that of a cloud.

for c) ii) the mean power INCREASES.
Since power = current x voltage
Just compare the graph you've drawn and the graph that's already given.
You don't need to consider the current cuz it stays constant.
What's the difference?
You see in the given graph the voltage drops to zero so the mean value is half of the peak value of voltage.
But your graph, the voltage never drops to zero so the mean must be way above half of peak value.
So the mean voltage or p.d. across resistor is increasing right?
And V is directly proportional to power.
So if voltage increases power must also increase.

 

[selrey]

in c)i) draw the smoothed voltage graph as shown in our book...in it the voltage never drops to zero.
It's shape is more like that of a cloud.

for c) ii) the mean power INCREASES.
Since power = current x voltage
Just compare the graph you've drawn and the graph that's already given.
You don't need to consider the current cuz it stays constant.
What's the difference?
You see in the given graph the voltage drops to zero so the mean value is half of the peak value of voltage.
But your graph, the voltage never drops to zero so the mean must be way above half of peak value.
So the mean voltage or p.d. across resistor is increasing right?
And V is directly proportional to power.
So if voltage increases power must also increase.

Thank youuu so much!
Also in part 1 the ms say that Vmin should be at 4 V. Like ik how the shape should be but how do we find the Vmin the graph should fall to?

 

[anastasia grey113]

Thank youuu so much!
Also in part 1 the ms say that Vmin should be at 4 V. Like ik how the shape should be but how do we find the Vmin the graph should fall to?

it says the variation is reduced to 1.6V.
So it should be like 5.6 to 4 to 5.6 to 4 again. The graph will have a variation of 5.6 -4 = 1.6V then.

 

[selrey]

it says the variation is reduced to 1.6V.
So it should be like 5.6 to 4 to 5.6 to 4 again. The graph will have a variation of 5.6 -4 = 1.6V then.

Ohhh right. Lmao i feel so dumb xD

 

[anastasia grey113]

Ohhh right. Lmao i feel so dumb xD

xD

 

[GarryTheGhost]

again with these graphs dont know how i should practice them :c, the only mistakes i'm making in the whole paper and like 70% worth of the paper is graphs lol.....
can someone explain why is it a negative gradient???? threshold frequency is the minimum frequency required to release an electron from the surface of a metal. is it negative gradient because the if wavelength increases then frequency decreases? and he ms also says concave curvature???? how would that look like???
also lol wtf do i do for the next part, i'm so oblivious in these types of questions.

 

[GarryTheGhost]

if the graph of displacement against time is starting from X0 i.e. the maximum displacement, then the one with coswt should be applied otherwise, sinwt should be used.

ahhh yes thought so, thank you for the clarification.

 

[anastasia grey113]

again with these graphs dont know how i should practice them :c, the only mistakes i'm making in the whole paper and like 70% worth of the paper is graphs lol.....
can someone explain why is it a negative gradient???? threshold frequency is the minimum frequency required to release an electron from the surface of a metal. is it negative gradient because the if wavelength increases then frequency increases? and he ms also says concave curvature???? how would that look like???
also lol wtf do i do for the next part, i'm so oblivious in these types of questions.

well just look at the equations.
For first graph...
E = hc/ λ so λ is inversely proportional to E so draw an inverse proportion graph.
For 2nd onw...
E = h/p so same inverse proportion.
the graph should be a curve with decreasing gradient.

 

[GarryTheGhost]

well just look at the equations.
For first graph...
E = hc/ λ so λ is inversely proportional to E so draw an inverse proportion graph.
For 2nd onw...
E = h/p so same inverse proportion.
the graph should be a curve with decreasing gradient.

assuming the first picture looks more like a curve.... is that correct?
also for the second part its the graph of
λ against momentum
since E proportional 1/λ
and E proportional 1/p
hence P directly proportional to λ hence a straight line???? Thats how i see it please correct lol.

 

[anastasia grey113]

assuming the first picture looks more like a curve.... is that correct?
also for the second part its the graph of
λ against momentum
since E proportional 1/λ
and E proportional 1/p
hence P directly proportional to λ hence a straight line???? Thats how i see it please correct lol.

well the equation that ik of is of De Broglie's wavelength and that's λ = h/p
i havent studied anything abt E's proportion to p

 

[GarryTheGhost]

well the equation that ik of is of De Broglie's wavelength and that's λ = h/p so

ah yes sorry my mistake sorry.
λ=h/mv
not E=h/mv

 

[GarryTheGhost]

well the equation that ik of is of De Broglie's wavelength and that's λ = h/p
i havent studied anything abt E's proportion to p

and obviously the second graph would be the 1/x curve hence not touching the axis since the denominator would never = 0
thank you very much for you help.

 

[anastasia grey113]

and obviously the second graph would be the 1/x curve hence not touching the axis since the denominator would never = 0
thank you very much for you help.

lol no problem yh true

 

[Holmes]

Can anyone tell me when to use which equation, like what are the situations or scenarios to use each equation. this is for SHM

look at C part

 

[Dtrickster]

View attachment 63418
View attachment 63419
Is it in our syllabus?
If yes then where should I find it's notes or other resource?
anastasia grey113 (the solution producer)

Idk if its in the syllabus but it has to do with transmission and receiving of waves(Waves)

 

[GarryTheGhost]

1. (a)= 2.6*10^-17N
   i need help in part (c) why did they multiply the the charge of point A BY 2? is it becuz they are also considering point B? but then why isn't the distance 12 micrometers instead of 6? since the separation is 12.
   how did they calculate the second line of the markscheme?