That makes sense, vanhee's first answer. I'm going to ask my tutor about this. It seemed like an error to me as well. It might be, since these questions I'm talking about were back in 1990's and the national science olympiads were not funded that much back then, so they weren't as professional...
Sure,
Question: A rectangular loop of wire with side lengths h and l is moved in the y+ direction in a magnetic field Bx=B/l (h-y), By=0, Bz=0. If the loops acceleration is a, find the emf generated as a function of time.
Solution's first line: Flux through the loop
Solution's other line...
Hello,
Short description: In my country's physics olympics, if a loop of wire is moved in a changing magnetic field, the time derivative of the area (while calculating emf with lenz law) is the velocity of the loop (as if we're taking the time derivative of the AREA SWEPT).
This has became an...
Two object are thrown at the same time from a surface which has an angle of θ. The first pbject is thrown parallel with the surface, with the speed v1. The second object is thrown horizontally with the speed of v2. The objects hit each other at a certain point. What is the distance between the...
Oh, right. So here I try again:
7 ( v + w ) = 3 ( 3 v + w )
2w=v
So the time taken when going with F
{\frac{140 (2w+w)}{2w-w}} = 420
And when going with 9F
{\frac{140 (2w+w)}{6w-w}} = 84
The answer is 5 thank you for your help :)
Ok I'll write it again.
140 (v + w) = 60 (9 v + w)
7 (v+w) = 3 (9 v + w)
7 v+7 w = 27 v + 3 w
4 w = 20 v
w= 5 v
But here, something goes wrong, while going from B to A, the speed is v minus w. And since w is 5 times v, it seems impossible.
If there's two velocities at the same direction, don't you add them up? If there's someone walking with 2 km/h in a train with 200 km/h, doesn't that make the person going with 202 km/h? Let the train represent the current and person represent the boat. That's why I add them up and call it the...
I think it's relative to the water, if you mean resistance by "water".
By V_{boat} I meant the speed caused only by the boat's engine.
Current is constant
And resistance is related with the boat's sum speed against the water squared
I don't know if I misunderstood the question, but I believe even if the boat haven't had an running engine, it should've the speed of current - resistance.
But I think there are 3 things that effect the boat, and one of them is towards the boat
The current
The boat's engine etc.
The resistance related to the boat's sum speed squaredSo,
Can the resistance be expressed like this: α (V_{current}+V_{boat})^2