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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. :/
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?
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?
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.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.
Thanks a lot guys.
Use vertical motion to find the time (time to travel vertical distance = time to travel horizontal distance)
Let me just make a last attemptI get it a bit
I guess the difference in angles between maximas should be same..
Why not B?
Look at the position of the ammeter. It's connected next to the fixed resistor and the fixed resistor is parallel to the variable one therefore the currents have to be different in both the loops i.e the ammeter is only giving you the current across the fixed resitance.
Why does it have to? :/
But then why doesn't the current through the fixed resistor change? Like the current is different in both loops
No! See you can't just change the resisitance of the fixed resistance if the current flowing through it is the same as it was before. Ratio of the resisitances does not increase in both of the loops. How can it? You shoud treat these loops seperately as seperate circuits. The main thing is the position of the ammeter, the resistance through it has to remain the same. If the ammeter was connected with both of the resisitors your theory would've been right but these are seperate two loops with seperate currents. Idk how to explain it. :/
Oh ok! So if the ammeter was positioned say between the two loops , then the reading would decrease?
I got it !!Let me just make a last attempt
Since it's travelling at 4m/s, the wave must travel, in 0.125s, 0.5m. Let's look at the wave step by step: (s is how much wave moved horizontally from initial position)
t=1, (s=4)
View attachment 54723
x hasn't moved yet. It's displacement = 0
t=1.125, s=4.5
View attachment 54724
x has moved down. It's displacement = -1
T=1.25, s=5
View attachment 54725
At this moment, displacement is still -1, but within this instance, it's gonna go all the way up.
T=1.375, s=5.5
View attachment 54726
x has moved up, displacement = +1
T=1.5, s=6
View attachment 54727
x is at +1 but is gonna move down in this very instance.
Now you may plot time against displacement. You'll see that it matches graph of B
That would be if all resistors were in parallel but there is a resistor in series too.
The weight of the man is supported by the 2 cables. Tension in 1 cable= (80*9.81)/2 = 392.4N
yes
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