Very helpful video! Thank you

Hi Jesse, fantastic video (as usual).

@jesseosbourne

2 жыл бұрын

Glad it was helpful! :)

Hey Jesse, or anyone in these comments! Any tips/hacks or tricks for solving antilogs with a number OTHER than 1? For ex: -log(2.5X10-4) or -log(7.1X10-4) I like the trick you used which blew my mind; counting the 'numbers' after the 1 to get an approximation; for ex: log(127) = approx 2, because if you count both 2 and 7, this is 2 numbers after the 1, therefore approx 2. HOWEVER, if the log isn't beginning with a 1 (ex: log(127)), and instead a 4 or 7, (for ex: log (403) or log(7992) how do we ensure it is still accurate using this method? Using your method Jesse, and continuing using the examples log(403). Would you then round up to log(1000), then count 3 numbers after the 1(that being three zeros), therefore we estimate the answer to be approx 3. Or roughly guestimate; log(403) there are two numbers after the 4, therefore (using both my trick and your trick) estimating it is between 2-3 applying both rules. Or for log(7992), similar method for solving; round up to log (10000), then there is 4 numbers after the log, so approx 4? Or again, applying both of our rules to solve, log(7992) there are 3 numbers after the 7 therefore the answer should be somewhere approximately between 3-4? Does my question make sense? Thank you all in advance for your time and efforts!

This if fantastic thankyou for breaking it down 😄

@jesseosbourne

2 жыл бұрын

No worries, glad it was helpful! :)

Hi Jesse, thanks for the video! I was just wondering for the last practice question, if we didn't have the dimensions of voltage already worked out, could we put voltage as IR, and therefore from R=V/I to R= IR/I? Or would we need to know the E= V I t formula?

@jesseosbourne

2 жыл бұрын

Hey Solange, that's a great question! So firstly, if you try to use the same formula to sub into itself you'll go round in a loop. So notice that by using V=IR to sub into R = V/I you get R = IR/I, the I then cancels out and you end up with R= R and a dead end. So this means that you do need an alternative formula when subbing in. In this question, I link from the previous question as a demonstration but I highly doubt that GAMSAT questions would actually be entirely dependent on answers from previous questions or else it would over rewarding and over penalising those that can and cannot do the first question. You could use E = VIt although I'd say it's less likely that they would require people to know quite so many formulas. The list that I run through at the start of the video is my prediction of what are likely expectations from ACER. That being said, the more formulas you know, the more options you have!

Hi Jesse, thanks for the video! I had trouble knowing which dimensions the variables were in e.g. acceleration is a measure of Length, where I thought it was Time. How do you approach understanding what dimensions a variable is? EDIT: Actually I think I just need to think more carefully about it, then I can do it :) Also, I put together a step by step process for Dimensional Analysis. I find it helps me to follow what's happening: STEPS 1) Rearrange to solve for the variable of interest 2) Replace variables with variables (from a formula) that have dimensions 3) Substitute all variables with dimensions 4) Simplify EXAMPLE F = Gm1m2/r2 1) Solve for G F = Gm1m2/r2 Fr2 = Gm1m2 Final: Fr2/m1m2 = G --> G = Fr2/m1m2 2) Replace variables with variables with dimensions F = ma = m(s/t^2) r2 = no change m1m2 = no change Final: G = m(s/t^2)/m1m2 3) Substitute all variables with dimensions G = ML/T^2L^2/M^2 4) Simplify G = M*L/T^2L^2/M = ML^3/M^2T^2 Final: G = L^3/MT^2

@jesseosbourne

2 жыл бұрын

Perfect, you've nailed it! I did also notice that in Q110 of the green book from ACER they snuck in a question that asked for the units as well which meant that you also had to identify equivalent units based on formulae (Nm = Force x displacement = Work = Joules) so certain questions require you to be flipping back and forth between use of formulae, units and dimensions too

@jaymie-leacollingwood2753

2 жыл бұрын

@@jesseosbourne Thanks Jesse! I’ll check that question out 😊. Also, do you recommend any specific Des questions or chapters?

@jesseosbourne

2 жыл бұрын

Truthfully I’m not super familiar with the des chapters yet as I’m still working my way through them in my own study but the chapter in visualising patterns has a lot of good organic chem practice in it It’s generally the volume and depth of the maths calculations that has des straying from ACER style questions in the other chapters

@jaymie-leacollingwood2753

2 жыл бұрын

@@jesseosbourne Thanks Jesse, that’s really helpful. I’ll continue skipping the super math heavy questions then haha

Hi Jesse, Have been finding the series very useful (haven't done maths for close to 10 years!) The scientific notation video in your Maths Series is on private, is there a reason for this or is it just an error? If it's unavailable, could you recommend a good learning resource instead? Thanks, really appreciate your content :)

@jesseosbourne

Жыл бұрын

Great to hear it's been helping, Niamh! It should be back up :)

Hi Jesse, are able to download the content you have created on Notion? There doesn't seem to be a duplicate option. No worries if you would prefer to keep it on the web only :)

@jesseosbourne

Жыл бұрын

Hi Jaymie, I have to leave it as web only just because of copyright protections

Hi Jesse, I'm watching this video with very little maths experience- last time i did any maths was 6 years ago so i was wondering if you could give me a definition/ explanation of why dimensions are used/ broken into and what they are? thanks :)

@jesseosbourne

Жыл бұрын

Hey Aimee, yeah sure. So it's not so much that quantities need to be broken into dimensions but instead it's that identifying the dimensions of a quantity can help understand what information it is giving you and how it can be applied mathematically to other quantities. The dimensions of a quantity are very much the same as dimensions of an object. Dimensions of a 2D shape are length and width and dimensions of a 3D object are length, width, and height. Really what this reveals is the way in which that quantity was calculated or what 'sub-quantities' go into making up its value. This is why dimensional analysis is so closely linked to use of common formulas. You'll notice in dimensional analysis though that length, width and height are all just categorised as a measure of length (L). This is a foundational dimension. The other common dimensions that aren't really broken down any further are mass (M), time (T) and sometimes Temperature (K) and Current (A or sometimes J) but these will always be identified in the question stem. The reason why we might need to understand dimensional analysis could be: 1. We want to understand its component dimensions ie. What is Force actually consisting of? F = ma and m is a dimension M, whilst a is a rate of change of distance with respect to time square (L T^-2) so putting this together Force has dimensions M L T^-2 2. We might also want to understand equivalence of dimensions or units. For example, using the above calculation, a newton has dimensions of M L T^-2. 3. We might want to understand the relationship between quantities and their base dimensions. This shows us that force is related is some aspect to the mass of the object, a distance, and the time in which it might cover this distance (inverse squared relationship) 4. Finally, we might just want to know if we can apply certain mathematical calculations to certain quantities. When adding quantities they must be of the same dimensions. ie. We can only add Force with Force because they share dimensions. I cannnot add a force in newtons and a length in metres because they do not have the same dimensions. I can however multiply and divide variables with different dimensions and possibly cancel out or combined dimensions. This is a very long winded answer for the sake of others reading and wondering the same thing but ultimately, GAMSAT boils this all down to a replicable skill as is shown in this video and my other dimensional analysis video. That being said, asking these questions is a good thing because it gives some context to what you're learning and why :)

@aimeesatur73

Жыл бұрын

@@jesseosbourne Thank you so much! I'm working through the maths material on your website and it's all starting to come together now

Hi Jesse, Juat wondering if we were not given V before question 3...what would the dimesnions of V be to work out R?

@jesseosbourne

10 ай бұрын

You'd have to determine it from other rules that calculate V from first principles (likely given in the stem). This was a series of exercises to simply demonstrate how derivation of dimensions work and in the case of Q3, how learnings from one part of a question can aid the solution in another.

Hi Jesse! I love your videos! You're amazing!! 😍😍 The explanations of dimensional analysis that you have provided are very clear and easy to follow! But I have a question, I noticed you used the formula F=G(m1m2/r2) and transposed the equation to find the dimensions of G. The final answer you got was G = L3/MT^2 (which makes sense). I stumbled across a Goldstandard textbook dimensional analysis question, which asked to find the dimensions of the SAME letter G from the SAME formula. But their final answer was G=M^-1L^3T^-2.... I was wondering why the Goldstandard textbook obtained a "different looking answer" to your answer, when they're BOTH finding the dimensions of G from the same formula. It's confusing me a bit now! Your answer makes more sense to me :). I don't get what Goldstandard even did! Why are they different? Sorry to bother you, or to take your time. May I please have this explained if that's okay with you? Thank you so much again for your videos! We are all forever grateful! 😀🥰🙏👍❤

@jesseosbourne

Жыл бұрын

Hey Jessica, wonderful to hear that the videos are helping so much! This one's actually a pretty simple fix. The solutions are equivalent but written in different formats. I've kept my solution in fractional form with positive indices while the Goldstandard solution has opted to write it out without the fraction and instead using some negative indices. Notice the denominator (M T^2) in mine has become M^-1T^-2 Order of the variables won't matter given they are all being multiplied. Hope this clears things up :)

@jessicakirjakovski8150

Жыл бұрын

Hello again :) I think I woke up this morning and realised that the answers were probably the same, only different format! But thank you so so much for confirming! 🙏😍🌺 My thinking process was: If you "split up the fraction" answer like so: L^3/MT^2 into L^3/1 × 1/M × 1/T^2 you can also perhaps see things more clearly - while making sure to align the doniminator with the denominator properly, when splitting things up etc. Now, all I did was plug in the negative index law's "other version" (M^-2 = 1/M^2) wherever I could, into the split up fraction: L^3/1 × 1/M × 1/T^2, as the "other version" number is the same thing and used interchangeably anyways. Hence, this means that the multiplication signs (from the split up fraction) will also remain the same. This makes sense, as the negative index law's "other version" we use (M^-2 = 1/M^2) are not even recipricals of each other, so shouldn't be treated as recipricals. It's literally just a matter of plugging in the "other version". Hence I got: L^3/1 × 1/M × 1/T^2 L^3 × M^-1 × T^-2 Final answers: L^3/MT^2 or L^3M^-1T^-2 Do you agree that this maths thinking process was all correct? Thank you once again! ❤😍🙏

@jesseosbourne

Жыл бұрын

Yep, all looks good!

hey jesse you said that acceleariotn is displacment over times squared. But why is displacement writen as S? Isn't S speed?

@jesseosbourne

2 жыл бұрын

Hey Levi! Yep so I use 's' for displacement because this is the variable generally used in most kinematics physics equations to represent displacement. You could use 'd' or 'x' as well but I've chosen 's' so that people can familiarise themselves with this fact as some past ACER questions have provided relevant kinematics equations (using s for displacement) that could be tricky for people who haven't caught this detail. it's just a bit annoying because yes, s is often used for speed as well. This is because speed is a scalar quantity (only has a magnitude) whereas the vector equivalent of speed (velocity) is represented using 'u' and 'v' in kinematics

Sorry another question😅pressure is included in the equations to memorise but i can't figure what dimensions would be its basic dimensions if that makes sense? Would it be mass?

@jesseosbourne

Жыл бұрын

So you have to start with P = F/A and then break these components down into their dimensions F = ma = M L T^-2 A = l x w = L^2 P = F/A = (M L T^-2) / L^2 P = M L^-1 T^-2

Not to be dramatic but you're my hero

@jesseosbourne

Жыл бұрын

Haha glad it's helped

hi jesse, i am confused, i thought velocity was distance/time = d/t, you said speed/time = s/t, did you mean to represent s as displacement and accidentally said speed?

@jesseosbourne

2 жыл бұрын

Hahah whoops! 🤦‍♂️ Yep, distance/time

@marcelitaa10

2 жыл бұрын

@@jesseosbourne thank you for clarifying :)

Hi Jesse, struggling to figure out why A/A = A^2, and T^2/T = T^3? Thanks

@jesseosbourne

2 жыл бұрын

Hey man, yep so this comes down to the order of operations and the importance of order in division. You'll notice that in these examples, the fraction has a fraction as a numerator and then a whole number denominator. ie. (a/b)/c This means that we cannot simplify b/c yet so instead we technically have to think of it as (a/b)/c = (a/b)/(c/1) = (a/b) x (1/c) = a/(bc) So you ca see that when dividing a fraction by a whole number, the whole number just multiplies to the denominator. This is why the power is increasing rather than cancelling out in these scenarios. To avoid the confusion between the fraction I usually write the main fraction line a lot larger than usual to 'dwarf' the size of my fractional numerator and/or denominator to help in seeing what I'm dealing with

@benjaminrichards7095

2 жыл бұрын

@@jesseosbourne Legend! Thank you