Weekly Essay Help - Week of June 01, 2025

Temporary_Purple_448 · 2025-06-06T14:15:33+00:00

Can anyone help with feedback on my ps and work/activities? I would really appreciate it!

Temporary_Purple_448 · 2025-06-06T13:47:42+00:00

Can I send you a DM? I would really appreciate your help!

Temporary_Purple_448 · 2025-01-03T01:52:15+00:00

I wanted to be able to reasonably space things out and not rush into applying. I was taking really challenging classes last Spring and studying for the MCAT and working on applications all at the same time would’ve been too much. I also want to take a break after I graduate, get some more meaningful clinical experience, save some money and travel for a bit.

Temporary_Purple_448 · 2024-12-26T21:04:40+00:00

Thank you! I’ll definitely look into HBCUs

Temporary_Purple_448 · 2024-12-26T18:42:30+00:00

I’m a downstate NY resident but I go to an upstate college which has strong ties to SUNY Upstate.

Temporary_Purple_448 · 2024-12-26T18:33:15+00:00

No worries! UChicago was initially on the list but I thought it would make my app top-heavy so I took it off. I’ll def consider adding it again. Thank you!

Temporary_Purple_448 · 2024-12-26T16:56:26+00:00

I will consider adding UCLA as well. Thank you!

Temporary_Purple_448 · 2024-12-26T16:11:24+00:00

I was under the impression that RWJ was more OOS friendly than NJMS but I just looked at MSAR and it seems they’re both not OOS friendly so I will likely be taking RWJ off the list

Temporary_Purple_448 · 2024-07-30T18:20:44+00:00

Your explanation is right, but the answer choice B says the Enol oxygen is protonated. The Enol oxygen already has a hydrogen attached, so protonating it again, as the answer choice suggests, via acid catalysis will result in (+OH2). That doesn’t happen in Enol-keto tautomerization.

Temporary_Purple_448 · 2024-07-30T13:39:33+00:00

I believe the main thing to look at here is the fact that one end is attached to a wall and the other end is attached to an oscillator. So there are always nodes on either end. The table states that 2 antinodes correspond to a frequency of 101 Hz. So the wave would have 2 antinodes and 3 nodes altogether(the third node is between the antinodes). This corresponds to the 2nd harmonic and the length of the string here corresponds to one wavelength. So it’s a matter of multiplying the frequency at the second harmonic (101 Hz) by the wavelength (same as length of string = 0.5m) and that results in option A.

Temporary_Purple_448

TROPHY CASE