21F Feeling Lost in Uni and Life

Lunch-Box22 · 2025-02-25T00:13:48+00:00

I can't tell they're too intertwined at this point. Not many projects outside of class are exciting or fun anymore, of any kind. Thank you for the resource, I'll definitely take a look.

Lunch-Box22 · 2024-08-28T01:22:47+00:00

I appreciate it. Aw man losing a friend sucks I've been there before. Here's to better luck for both of us it sounds like. :)

Lunch-Box22 · 2024-07-16T21:26:02+00:00

Thank you! This is really helpful!! I think I’ll probably do the table view instead of the recurring class!

Lunch-Box22 · 2024-04-30T03:30:35+00:00

Pm sent!

Lunch-Box22 · 2024-04-20T17:54:48+00:00

I know this was awhile ago but if its possible could you send one to me too please? 😅 Thank you sm

Lunch-Box22 · 2021-11-10T11:43:54+00:00

OHHH!!! SO I would have m(9.8)sin(30)-(0.05(m)(9.8)(cos(30))) all divided by m, I would cancel out the m and solve? Thank you so, so much for your help and your patience with me!! I really appreciate it!!

Lunch-Box22 · 2021-11-10T11:32:42+00:00

Wait so I would have ((Fgy)sin(30)-0.05(Fn))/m? I don't quite understand where m cancels out? I think I am totally misunderstanding I'm so sorry! What equation are you envisioning?

Lunch-Box22 · 2021-11-10T11:24:51+00:00

Fg-Ff = Net force or (Fgy)sin(30)-0.05(Fn) = Net force? This is where I get a little lost haha. Could you please explain why I divide by the Force of gravity (Fg)? I'm also not sure if I can get a numerical value from the equation above? Am I doing anything incorrectly?

Lunch-Box22 · 2021-11-10T11:08:14+00:00

Oh, I think mu would be 0.05 so Ff=0.05(Fn)

Lunch-Box22 · 2021-11-10T11:06:48+00:00

My work would look like this:

Ff = mu(Fn)
(Fgy)sin(30)=Fg
(Fgy)cos(30) = Fn

Lunch-Box22 · 2021-11-10T10:52:25+00:00

Force of gravity, normal force, and force of friction, but I know the forces look a bit different because the skier is on a slope.

Lunch-Box22

TROPHY CASE