I guess the v*sec(α) and then due to energy conservation I get the result. I think I got it now though it`s not easy to understand.
Thanks for the help!
If I understand correctly than with velocity v. I still not undertand what you mean (not because you expressed it badly, I just don`t see the connection). If you could explain in more details how it might lead to the solution I should fully understand it. Sorry if I don`t see something obvious...
If the molecule is stationary then with velocity v. If it has a "sideways" velocity then the square root of v2 and v`2 where v` is the sideways velocity. But I`m not sure what you mean by this.
Energy concervation tells me that the velocity is constant since there is no pressure drop or change in height. So after collision water will move upwards with velocity v. I still don` understand why it`s not the solution.
I was trying to take into account that new `layers` of water hit the...
Is the velocity unchanged due to energy conservation? If so I can`t figure out the solution. I also don`t know why would it change. Maybe I mess something up with the geometry.
Yes, however in th part b the answer is u=v/cosα where α is the angle with the vertical, u is the velocity of the water upwards. So there the velocity changes hence applying the energy concervation is`nt as straightforward.
Hi,
I`m seeking for help in the following problem.
A flat vertical board is traveling in water which is to be considered as ideal. One of its ends is in water, the other one is outside the water. Its velocity is v with respect to its normal. What is the velocity of the water stream directed up...
https://www.ioc.ee/~kalda/ipho/meh_ENG2.pdf
At the 21st page problem 52 is basically the same, after the problem there are hints, but I don't understand them
The solution is 2g
First notice that the only horizontal force is the force F, so the bar will have horizontal acceleration. If you choose the pivot point somewhere then it's like you choose the frame of reference of that point, which accelerates.
Now if you choose an accelerating frame of reference then you have...
Hi,
I have a question about a rising bubble.
I read that the initial acceleration of a bubble (with negligible mass) in water is 2g, where g is the gravitational acceleration. I understand that if a bubble rise then the water move around it, but I can't derive this equation.
Could someone help...
Have you drawn the normal force N?
You should write out the balance of vertical forces to find N.
Then choose the centre of mass as the pivot point of torques and write out the torque balance of the force F and N.