Recent content by Lui4

I don't have a clarifying answer. We are just given these questions as homework. So Im not too sure. I would assume its 10cm below the water regardless of the oil or not Update. The correct answer was 241 but how?? Like do we just not consider before the oil was added. since to get 241 as the...

yeah I didn't mean to do that but I knew that number is correct. Could you please tell me how to calculate the pressure after adding the oil? Because then you just take the difference of those two and I think the problem is solved.

I appreciate your efforts and all the time you took to write that out and make sure it was nicely written in a way that's easy to understand and all those units you put. But for whatever reason that is not the correct answer. Thank you for trying to be of help, its greatly appreciated!

No worries and okay sure I will be patiently waiting for your response, thank you!

Yeah this is what I did too but the only difference is, if this is before the oil then we wouldn't add the 0.03m right? since that is the height contributing to when the oil is added?

Yeah I am assuming that data is the area value they gave us in this case

so we are told that we want the pressure below the water at 10cm for before and after so we know h and g is a constant and p we also know since its the density of the water so when you multiple these values that's where the 980 comes from no? ( 0.1 * 1000 * 9.8)= 980

I think that's pretty self-explanatory. I understand what to do I just don't see how I am supposed to do it and that's why I am asking for help on the calculation aspect of it. Lets start with the first thing you said. Lets find the pressure before. Okay great, now this is without the oil being...

okay yeah I see I've made a couple mistakes. If my work doesn't make sense, sure, fair. But could you then instead comment on what I should be looking to do?

Because I understood that we have two pressure points and they both have different heights and that is why i tried separating the two apart and I used Patm for the oil since that is what the atmospheric pressure is going to be applied to since I'd assume the u tube is open from the side the oil...

oh thats a good point. I drew a little u tube but I don't see what to do from here. Because in this case we are only focusing on one side of the tube anyways so it wouldn't matter about the other side since they ask us about the pressure of only one side where the oil was added no? If you can...

Mark Numbers? We need numbers for the x and y-axis and that's what I'm trying to figure out. After we figure that out I don't know what to plot.

Okay I understand the point you made of if the skier was in free fall it would be 9.8 but since he is moving down a slope It's not 9.8. I also understand that the skier moves at a constant rate. The last part of your response is where I'm confused. My only guess would be 1/2 is equal to 0.5...

Yes this does ring a bell this is taken from the formula d=vot+1/2at^2. Now that I know this formula what exactly do I need to do with it as there is no numbers given. The solution that I think using this equation would be 0.5 x 9.8 which would leave us at D=4.9t^2. I don't think that is correct...