Physics: Post your doubts here!

[qwertypoiu]

Thanks a lot ^.^

Btw,the ans to that ques is B ( I thought D too).. I'm attaching the screenshot of Examiner report .. see if it makes sense to you if it does then explain me too

Ah I misread that question. I thought it was asking what the displacement graph would look like 4 seconds later. But rather, what they're asking is about how the point X on the far right of the diagram given would move over the course of the next 4 seconds. We know that the wave will travel 4 metres in 1 second.
At t=1:

The pulse will move like this. This explains why the point X remains stationary from t=0 to t=1.
As the wave continues traveling to the right now, you must imagine how X will move. It will move down, then up all the way, then back down. I'm sorry if this didn't make sense but I don't know how to explain the moving image I have in my mind
This leaves only A and B as choices. A is moving too slow, by t=2, the pulse would have long disappeared on the right.

 

[qwertypoiu]

9702/11/O/N/10
Q34

 

[Boss201]

ANS : C how ?

 

[princess Anu]

Ah I misread that question. I thought it was asking what the displacement graph would look like 4 seconds later. But rather, what they're asking is about how the point X on the far right of the diagram given would move over the course of the next 4 seconds. We know that the wave will travel 4 metres in 1 second.
At t=1:
View attachment 54703
The pulse will move like this. This explains why the point X remains stationary from t=0 to t=1.
As the wave continues traveling to the right now, you must imagine how X will move. It will move down, then up all the way, then back down. I'm sorry if this didn't make sense but I don't know how to explain the moving image I have in my mind
This leaves only A and B as choices. A is moving too slow, by t=2, the pulse would have long disappeared on the right.

I get it a bit, What I still don't get is why wud the pattern be like up down down up.. but it s okay thanks a lot !

 

[princess Anu]

9702/11/O/N/10
Q34
View attachment 54704

 

[eliyeap]

9702/11/O/N/10
Q34
View attachment 54704

First find the resistance of the 2 wires: Resistance is 0.005 ohm per m, the length of a wire is 800m
R for 1 wire= 800*0.005=4 ohm
Total R for cable (2 wires)= 4*2=8
Current in circuit= 0.6A
PD across cable= 0.6*8=4.8
Emf= 4.8+16=20.4V

 

[eliyeap]

View attachment 54705
ANS : C how ?

WD by the Driving Force= PE gain + KE gain
Here question says minimum WD, they give us no information on the speed whatsoever, so assume that there is no KE gain
WD = PE gain= 50*9.81*1.6= 784.8J = 780J

 

[eliyeap]

How??

Resolve the Tension of the string perpendicular to the weight= Tcos30=0.15 T=0.1732N
Now resolve the tension of the string horizontal to the air flow= 0.1732* sin30= 0.087N

 

[69ermalik007]

Help.....

 

[69ermalik007]

 

[69ermalik007]

 

[69ermalik007]

 

[69ermalik007]

 

[69ermalik007]

 

[Mahnoorfatima]

Help.....
View attachment 54709

Look the question has itself given you the answer "constant velocity" What does this mean? Forces are at equilibrium or are balanced thus look for such a diagram which would follow the course of a complete triangle. Look at the arrows. Every arrow should lead to a complete triangle. Idk how to explain it but thats the best I could do. :/

 

[nehaoscar]

The reason of D + it shouldn't be I/R, IR is correct.

Upload a file then full image.

But in parallel the formula is 1/R = 1/R1 + 1/R2 etc... so if you do I/(1/R1 + 1/R2 ...) then it will give you IR right?

 

[nehaoscar]

Current is unchanged because it's connected with the fixed resistor which will have a different reading as compared to the one with the variable one as they're connected in parallel and the reading of the voltmeter decreases because there's an increase in resisitance leading to a decrease in the current in this part of the circuit. thus reducing the Pd as well.

Please could you explain why the current is unchaged though as I don't understand that one? Like the total resistance of the circuit decreases, so the current increases. The ratio of the resistances at the junction would increase so more current would flow to the fixed resistor at the top so shouldn't ammeter reading increase?
But then why is it unchanged?

 

[nehaoscar]

.
Why not B?

 

[nehaoscar]

Just asking, How do you guys post the screenshot thingy? I don't usually get answers if I post the whole paper and ask for a certain question.

Either use the snipping tool if you have on your laptop. Using that it automatically copies the image that you have cropped and then past it on here.
Or if you don't have the tool then just take a screenshot of the question you want and paste it here

 

[qwertypoiu]

First find the resistance of the 2 wires: Resistance is 0.005 ohm per m, the length of a wire is 800m
R for 1 wire= 800*0.005=4 ohm
Total R for cable (2 wires)= 4*2=8
Current in circuit= 0.6A
PD across cable= 0.6*8=4.8
Emf= 4.8+16=20.4V

Thanks a lot guys.