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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.
Someone please explain why C??
The reason of D + it shouldn't be I/R, IR is correct.
Upload a file then full image.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.
It is in parallel with the other 3 resistors so it will have the same p.d. as the series combination.Four resistors of resistance R, 2R, 3R and 4R are connected to form network.
Battery of negligible internal resistance and voltmeter are connected to the resistor network as shown.
Voltmeter reading is 2 V.
What is electromotive force (e.m.f.) of the battery?
A 2 V B 4 V C 6 V D 10 V
The solution given on that physics website goes like:
Ohm’s law: V = IR
At the 2R resistor, the voltmeter reading is 2V. So, current I through it = 1 A and current I is constant through any specific loop (since the resistors are connected in series).
So, through the 2nd loop, total resistance = 3 + 2 + 1 = 6R
e.m.f. = total p.d. = IR = 1 x 6 = 6V
They've ignored the 4R which is parallel to the other three R's. WHY?????
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.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
I get it a bitAh 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.
, What I still don't get is why wud the pattern be like up down down up.. but it s okay thanks a lot !
First find the resistance of the 2 wires: Resistance is 0.005 ohm per m, the length of a wire is 800m
WD by the Driving Force= PE gain + KE gain
Resolve the Tension of the string perpendicular to the weight= Tcos30=0.15 T=0.1732N
How??
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