Physics: Post your doubts here!

[Mush999]

Precision is the number of decimal places to which a measurement is taken, while accuracy is the closeness of the measurements to the actual values. So 2.33 is more accurate(closer to 2.321, you can find this out by subtracting the measurement and the actual value), while 2.344 is more precise since it's measured to a greater number of decimal places. This makes C the answer.

Thank you so much brother!

 

[A*****]

Does frequency need to be the same for two waves to be coherent?

 

[hellodjfos;s'ff]

Both of them please

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For 36, you have to apply kirchhoff's 2nd law. There are two points where the voltmeter is connected, I will name the point on the right, A and the one on the left, B. So at A, the pd is E because it just came out of the cell and hasn't been used up(I am treating the internal resistances as resistor so that it's easier). This means that the pd at B must also be E since the potential difference is zero between these points. This means that the E at A was used up by the r1 since the more pd(also E) came out of the new cell. So now we can form 2 equations. First one: E=Ir1(just using V=IR), second: 2E=E+I(r2+R) which simplifies to become: E=I(r2+R). Now I can solve them simultaneously. I replace the E in E=I(r2+R) with Ir1(which is my other equation, E=Ir1).
This becomes Ir1=I(r2+R) ----> r1=r2+R ----> r1-r2=R making B the answer.

 

[hellodjfos;s'ff]

Thank you so much brother!

No probs bro

 

[hellodjfos;s'ff]

Does frequency need to be the same for two waves to be coherent?

Not necessarily, if waves need to be coherent, they must have a constant phase difference. Usually when waves are produced to be coherent they use the same source to produce the waves so yes, mostly they have the same frequency but it's not required.

 

[A*****]

Two wires with the same Young modulus E and cross-sectional area A, but different lengths L,
are subject to different tensile forces F. The extension e of each wire is the same.
The column headings in the table show four different quantities.
Which quantities have the same value and which quantities have different values for the two
wires?

FL/e. Ae/L. E/FL
A different differen same
B different same same
C same different different
D same different same

If u rearrange the equation E= Fl/Ae by making Fl/e the subject, it would be Fl/e= EA, since E and A are same for both so Fl/e would also be the same
Next make Ae/l the subject, Ae/l= F/E... E is same but F is different for both so F/E and hence Ae/l would be different
E/Fl= 1/ Ae A and e are same so E/Fl would also be same

 

[A*****]

For 36, you have to apply kirchhoff's 2nd law. There are two points where the voltmeter is connected, I will name the point on the right, A and the one on the left, B. So at A, the pd is E because it just came out of the cell and hasn't been used up(I am treating the internal resistances as resistor so that it's easier). This means that the pd at B must also be E since the potential difference is zero between these points. This means that the E at A was used up by the r1 since the more pd(also E) came out of the new cell. So now we can form 2 equations. First one: E=Ir1(just using V=IR), second: 2E=E+I(r2+R) which simplifies to become: E=I(r2+R). Now I can solve them simultaneously. I replace the E in E=I(r2+R) with Ir1(which is my other equation, E=Ir1).
This becomes Ir1=I(r2+R) ----> r1=r2+R ----> r1-r2=R making B the answer.

Could it also be r2-r1 if it was in the options?? Bqz the same equation can be formed with r2

 

[hellodjfos;s'ff]

Both of them please

View attachment 64061

For 37, again you apply kirchhoff's 2nd law. Two cells are supporting each other while one is opposing. 2E-E=I(R+R+R) ---> E=I(3R) ----> R=E/3I . Now, you can find the pd used up by resistor in the left part of the circuit using V=IR. V=I * E/3I ----> V=E/3 . So that means that a remaining 2E/3 pd goes to P. Now since the cells in the upper and right part of the circuit are opposing each other, their emfs are cancelled so the pd at Q becomes 0. So the potential difference between P and Q is 2E/3 - 0 making the answer 2E/3(C).

 

[A*****]

For 37, again you apply kirchhoff's 2nd law. Two cells are supporting each other while one is opposing. 2E-E=I(R+R+R) ---> E=I(3R) ----> R=E/3I . Now, you can find the pd used up by resistor in the left part of the circuit using V=IR. V=I * E/3I ----> V=E/3 . So that means that a remaining 2E/3 pd goes to P. Now since the cells in the upper and right part of the circuit are opposing each other, their emfs are cancelled so the pd at Q becomes 0. So the potential difference between P and Q is 2E/3 - 0 making the answer 2E/3(C).

Thank u sooooo muchhh!!

 

[hellodjfos;s'ff]

Could it also be r2-r1 if it was in the options?? Bqz the same equation can be formed with r2

If you work the other way around, I think you should get the same equation, regardless of that, it wouldn't matter whether it's r2-r1 or r1-r2.

 

[hellodjfos;s'ff]

Thank u sooooo muchhh!!

Your welcome

 

[A*****]

This may sound silly but hope u understand my question
This is the diffraction pattern right?? What if we had a double slit and there was an interference pattern...would it also be a horizontal pattern? Like when we draw diagrams we make the pattern as if it was vertical

 

[Kanekii]

Need help with these

 

[A*****]

Shouldn't D be correct here as the air resistance is negligible and using the equation s=ut + 0.5at² isn't it possible?

 

[A*****]

Need help with theseView attachment 64063 View attachment 64064 View attachment 64065 View attachment 64066

For 11, 2(1) +3(-4) = 5x
x= -2 and I used -ve for left side so it is 2m/s to the left

 

[A*****]

Need help with theseView attachment 64063 View attachment 64064 View attachment 64065 View attachment 64066

 

[aisya farooq]

how to solve this???

 

[hellodjfos;s'ff]

This may sound silly but hope u understand my question
This is the diffraction pattern right?? What if we had a double slit and there was an interference pattern...would it also be a horizontal pattern? Like when we draw diagrams we make the pattern as if it was vertical

Yes, there would be diffraction in this question. Yes, if you had a double slit, you would also have a horizontal pattern. In diagrams, the pattern appears to be vertical but in reality it's horizontal. If you do a practical for this you would know(CIE encourages this).

 

[hellodjfos;s'ff]

Yes, there would be diffraction in this question. Yes, if you had a double slit, you would also have a horizontal pattern. In diagrams, the pattern appears to be vertical but in reality it's horizontal. If you do a practical for this you would know(CIE encourages this).

Also, search for images of actual experiment results for double slit and diffraction grating to understand this. Both are horizontal. The diffraction grating spreads the light into the colour spectrum, horizontally. While, double slit produced dark and bright fringes horizontally as well.

 

[hellodjfos;s'ff]

Need help with theseView attachment 64063 View attachment 64064 View attachment 64065 View attachment 64066

For 9, since the force of gravity acting on her is constant, her acceleration and deceleration would be constant too. Consequently, the gradient of the graphs must be constant too, this rules out B and C. To decide between A and D, find the diagram with a completely vertical line which would be D. There would be a completely vertical line because when the girl is on the trampoline, the spring of the trampoline suddenly causes her velocity to change direction(go opposite) but remain of the same magnitude. This means that her velocity changes instantly(no time taken). So the answer is D.

A*** already explained 11 and 24.

For 26, you find the period of the wave using T=1/f . The period will turn out to be 4 ms. Now in a stationary wave, the wave goes up(max displacement), zero displacment, down(minimum displacement). So using this concept, for this wave, the wave will go to zero displacement in 1st ms, then up in the 2nd ms, then zero displacement in the 3rd ms, then down to the original position in the 4th ms, and then back to zero displacement in the 5th ms. That means after 5ms, the wave will have zero displacement giving the answer B.