Physics: Post your doubts here!

[anastasia grey113]

View attachment 63388
oct nov 2017 13
any help would be appreciated

i think it should be D.
Can't be B or C cuz of Newton's 3rd law so the force by floor on man must always be equal to the force on floor by man.
However, since the floor itself is accelerating downwards, it must be applying lesser force on the man than a still floor.
So the force must be lesser than the weight of the man.

 

[ErosKuikel]

Does anyone have mark scheme definition list for Alevel physics? Plz share

 

[ErosKuikel]

Hi guys, this is kinda unrelated to physics but I wanted some help with further maths question since further thread is pretty much dead. If anyone taking further maths can help me out it would be very helpful. I can do (a) but I need help with (b). It's S07P1Q11 btw if you need Mark Scheme which didn't help me enough to understand.

Hope it helps!

 

[Minsa Fatima]

For J08 P5 Q2 d why won't we include s^-1 in the units

 

[GarryTheGhost]

THESE GRAPH QUESTION ARE SO ANNOYING CAN SOMEONE HELP ME ON THESE just the graph parts of the questions, also for the rectifier question where would P and the capacitors go?
thanks in advance

 

[GarryTheGhost]

For J08 P5 Q2 d why won't we include s^-1 in the units

R=R0e^-pnx
(R/R0)=e^pnx
Ln(R/R0) divided by px= n
since the units of R and Ro is 1/s when doing R/R0 you are basically doing 1/s divided by 1/s which is the same as 1/s multiplied by s/1 hence the s cancels out leaving you with:
1/px=n
units of p=kgm^-3 and x=m
since its kgm^-3 the m^3 goes up and it cancels with the m at the bottom leaving you with units of n as m^2Kg^-1

 

[anastasia grey113]

THESE GRAPH QUESTION ARE SO ANNOYING CAN SOMEONE HELP ME ON THESE just the graph parts of the questions, also for the rectifier question where would P and the capacitors go?
thanks in advance

1-
Since this is a non inverting amplifier you have to use the following formula.
Gain = Rf/Rin + 1
2- Now Vout is found by multipilying Vin x gain so do this for all values for the graph. The graph will be a straight line but don't stretch it above 9V cuz you can't have a Vout above 9V. At that point, the Vout stay 9 no matter how much u increase the Vin. Because the op amp is 'saturated'.
3 and 4- The current is directly proportional to change in flux but EMF is directly proportional to RATE OF CHANGE of flux. So take the gradient. It will be constant and positive initially till t1, then constant and negative (more negative) till t2 and then it becomes zero.
5 and 6 -
The P is shown below marked red. You can trace the path of the current passing thru the diodes (always passes thru diodes having the P junction before). Current always flows from positive to negative terminal of the load.

For capacitor, put it in parallel to the load like this. This is how it's placed in every rectifier for smoothing: in parallel to the load.

 

[anastasia grey113]

GarryTheGhost thats how v draw em


THE 2ND ONE IS NOT THAT ACCURATE BUT JUST TO GIVE U AN IDEA

 

[Holmes]

Terrifying graphs.

Can anyone tell me how to draw a worst fit line if error bars lies on the x axis june 2017 52: reference
anastasia grey113 can you help me?

 

[anastasia grey113]

Holmes
yh sure lol
the diagram below is just for explanation I'm running short of time..i apologise for the not-so-nice quality of the graph lol
okay so start with drawing the error bars along the x axis (shown by blue)
the two error bars ive drawn here are of the first and the last point which coincidentally lie on the line
the worst acceptable line (red) is the one which passes thru the extreme left of the error bar of the last point and the extreme right of the error bar of the first point and vice versa.

 

[Holmes]

Holmes
yh sure lol
the diagram below is just for explanation I'm running short of time..i apologise for the not-so-nice quality of the graph lol
okay so start with drawing the error bars along the x axis (shown by blue)
the two error bars ive drawn here are of the first and the last point which coincidentally lie on the line
the worst acceptable line (red) is the one which passes thru the extreme left of the error bar of the last point and the extreme right of the error bar of the first point and vice versa.
View attachment 63404

Thanks. It was a HUGE help.
Thanks again.

 

[hamadh]

View attachment 63318
Can anyone help me with this ques.? Explanation needed..☺

The current in the horizontal resistors is 8,3 and 1 respectively and across the vertical resistors is 5, 2 and 1 respectively. By this calculate voltage across each resistor and the total potential difference across the 1st horizonal resistor and 1 vertical resistor gives us the voltage across the battery's input terminal. Now the thing that is of our main concern : Current. As you can see in the last section where the current is 1 A the total voltage across the two resistor is 2V . These two resistors are parallel to the second vertical resistor so it is supposed to have 2V across it too as branches have same voltage. By this the current in second branch is 2 A . By kirchoff's first law, the current flowing through second horizontal resistor will be 3 A so voltage across it will be 3V. Now the 2nd horizontal resistor and the second vertical resistor are in parallel to first vertical resistor so it has a voltage of 5V . By this, we know that current across it is 5A . So the total current flowing through the first horizontal resistor is 8A.

 

[hamadh]

View attachment 63388
oct nov 2017 13
any help would be appreciated

As the lift is accelerating down, there is a resultant force acting on the man and the lift (2nd law of newton)
F=ma the resultant force F is equal to weight of the man minus the tension in the cable of the lift so technically the force exerted by the man us less that its weight.

 

[Holmes]

Help needed in b(ii)
anastasia grey113
Is there any specific formula for this???

 

[anastasia grey113]

Holmes
it says it samples at a frequency of 44.1kHz so that means it takes 44.1 samples in 1 second and each sample is of 16 bts.
So for 1 second we have =16 x 44100 = 705600 bits
And for 5 mins 40 s = 340 seconds, we hav = 705600 x 340 = 2.4 x 10^9 bits

 

[Holmes]

Holmes
it says it samples at a frequency of 44.1kHz so that means it takes 44.1 samples in 1 second and each sample is of 16 bts.
So for 1 second we have =16 x 44100 = 705600 bits
And for 5 mins 40 s = 340 seconds, we hav = 705600 x 340 = 2.4 x 10^9 bits

thx

 

[anastasia grey113]

Q6 part c)
I viewed the mark scheme but i still don't get it

 

[Gouki_Destruction]

View attachment 63417
Q6 part c)
I viewed the mark scheme but i still don't get it

Im still an Olevels student , but my knowledge suggests its bcz photons are said to b massless , or as small as 1X10^-18 , while an electron weighs approx. 1/1840 of a proton . Considering this theory , and equating it by the formula mv-mu , the change in momentum is found . Naturally , if the masses are different, the momentum change has to different too , thus taht is my suggestion , probably wrong i guess

 

[Hamnah Zahoor]

36)
Applying KVL around the circuit gives
2E = IR + Ir1 + Ir2

Since the voltmeter reading zero this means there is no voltage difference between its contacts points which gives
Vrise=Vdrop
E = Ir1

Substituting this into the first equation gives
2Ir1 = IR + Ir1+ Ir2
R = r1 - r2 (B)

37)
Applying KVL around the circuit gives
2E - E = 3IR
E = 3IR

The PD between P and Q is due to the battery and resistor between them
PD = E - IR
= E - 1/3 E (substituiting first equation)
= 2/3 E (C)

I would recommend practicing how to use KCL and KVL using YouTube videos.

Thankyou.

 

[anastasia grey113]

Im still an Olevels student , but my knowledge suggests its bcz photons are said to b massless , or as small as 1X10^-18 , while an electron weighs approx. 1/1840 of a proton . Considering this theory , and equating it by the formula mv-mu , the change in momentum is found . Naturally , if the masses are different, the momentum change has to different too , thus taht is my suggestion , probably wrong i guess

lol idk i have this feeling that u can explain it to me
here's wut the mark scheme says
(c) momentum is a vector quantity B1
either must consider momentum in two directions or direction changes so cannot just consider magnitude B1 [2]
u got any idea wut this means?