Hey Jesse, thanks for your content, it is such a huge help. I just had a query regarding the last question: you wrote that alpha was equal to M.L-2.T-2, but the whole bracket contained K/V + 2alpha - so why would those dimensions not be M.L.T-2 instead due to K/V = alpha. Sorry if its obvious, I am new to dimensional analysis and math stuff. Thanks heaps
Hey Alex, glad it's helping! Hmm you know that might just be an error on my part because I was making up fake formulas haha. You're absolutely right that you could solve alpha as MLT^-2 because if K/V = alpha in dimensions then since K/V+alpha = F, the dimensions of alpha should also be equal to that of F which is MLT-2. If you're interested, I may as well explain my assumptions and why they were wrong because there's some good mathematical reasoning that could be handy. I assumed when making this video that if I simply make up equations and solve for one of the variables and its dimensions, it should always come out consistently because I'm solving one variable from one equation. BUT, what I didn't think about was the fact that the equation setups can come from multiple different rules relating to dimensions (the very ones I was trying to explain and demonstrate sequentially in this video, ironically!). Because we could use the rules such as ' if (a + b) exists then a and b must have the same dimensions' and 'if LHS = RHS' then the dimensions of the LHS and RHS must be equal' and 'if ab = c then the product of the dimensions of a and b must be equal to that of c', THEN, there are actually multiple equations that could be created and so I had actually created a simultaneous equations problem! This meant that in the video, I unknowingly was only solving using one method assuming regardless of which one I chose to demonstrate in the example, it would give me the same result but it was actually spitting out a result based on that method and different ones for other methods. Hopefully that makes sense, but it's also not entirely necessary, I just really enjoy this stuff. All in all, yes, your reasoning is right. The question is flawed because of the fact that it is made up. If using actual real equations (which ACER always will), then it won't matter which dimension rule you use to set up your equation, it'll give a consistent answer. I'd say you're on a good path with your dimensional analysis if you caught that error and queried it so good stuff!
Hi Alex I thought you can't take the answer from Q2 (K's dimension) and equate that dimension to Alpha, because you actually need to equate K/V dimension to alpha, not equate K dimension to Alpha. In the original formula you are adding (K/V) + alpha, not K + Alpha. I wonder if that is why?
Hey Jesse! Thanks for the video! Can you go through the 3D graphical analysis and questions? Especially with state changes such as solid, liquid and gas. Thanks!
Hi Jesse, i am having a hard time understanding something - why would the dimensions of the second bracket not just be Amps? Seeing as we worked out that Q = amps and the dimension of anything added/subtracted in one expression must all be the same dimension?
Hi Stella, so Q was charge which was equal to zt but Q is not the same as current. This proof is given by the maths and the use of the formula Q = It where I = current in amps because zt is subtracted from Q, the dimensions of Q and zt must be equal, ie. Q = zt Using the formula Q = It, we can replace Q with It to give It = zt and then t can cancel out from both sides giving I = z so the dimensions of z are amps. Hopefully that clears things up!
Hey Jesse, thanks for the video, really helpful. Just at 6:48 I'm struggling to see how you went from M.L.(T^2)/L^2 to M.(L^-1).(T^-2)/L. Wouldn't the L on the top be cancelled out and your just left with L^1 on the bottom. So instead of the final value of P=M.(L^-2).(T^-2), shouldn't it equal P = M.(L^-1).(T^-2)? Have I misunderstood something?
Hey, I'm not Jesse ahaha but I'll try my best to explain! So we calculated (M.L.T^-2)/L^2 as the dimensions for Pressure. Just like you said, the L on the top will be cancelled out by the L^2 on the bottom. So a simplified version of this equation will be (M.L^-1.T^-2) which equals Pressure which is what you got too! Jesse then calculates the dimensions for the alpha (a) constant which has the units of Pressure/Metres. So here we plug in our dimensions for Pressure (M.L^-1.T^-2) and use L as the dimensions for Metres. The final dimensional equation for (a) would then be (M.L^-1.T^-2)/L which is then simplified to (M.L^-2.T^-2) - this is the equation Jesse ends up with at the end. Hope that helps!
Hey Jesse, thanks for your content, it is such a huge help. I just had a query regarding the last question: you wrote that alpha was equal to M.L-2.T-2, but the whole bracket contained K/V + 2alpha - so why would those dimensions not be M.L.T-2 instead due to K/V = alpha. Sorry if its obvious, I am new to dimensional analysis and math stuff. Thanks heaps
Hey Alex, glad it's helping! Hmm you know that might just be an error on my part because I was making up fake formulas haha. You're absolutely right that you could solve alpha as MLT^-2 because if K/V = alpha in dimensions then since K/V+alpha = F, the dimensions of alpha should also be equal to that of F which is MLT-2.
If you're interested, I may as well explain my assumptions and why they were wrong because there's some good mathematical reasoning that could be handy.
I assumed when making this video that if I simply make up equations and solve for one of the variables and its dimensions, it should always come out consistently because I'm solving one variable from one equation. BUT, what I didn't think about was the fact that the equation setups can come from multiple different rules relating to dimensions (the very ones I was trying to explain and demonstrate sequentially in this video, ironically!). Because we could use the rules such as ' if (a + b) exists then a and b must have the same dimensions' and 'if LHS = RHS' then the dimensions of the LHS and RHS must be equal' and 'if ab = c then the product of the dimensions of a and b must be equal to that of c', THEN, there are actually multiple equations that could be created and so I had actually created a simultaneous equations problem! This meant that in the video, I unknowingly was only solving using one method assuming regardless of which one I chose to demonstrate in the example, it would give me the same result but it was actually spitting out a result based on that method and different ones for other methods. Hopefully that makes sense, but it's also not entirely necessary, I just really enjoy this stuff.
All in all, yes, your reasoning is right. The question is flawed because of the fact that it is made up. If using actual real equations (which ACER always will), then it won't matter which dimension rule you use to set up your equation, it'll give a consistent answer. I'd say you're on a good path with your dimensional analysis if you caught that error and queried it so good stuff!
Hi Alex I thought you can't take the answer from Q2 (K's dimension) and equate that dimension to Alpha, because you actually need to equate K/V dimension to alpha, not equate K dimension to Alpha. In the original formula you are adding (K/V) + alpha, not K + Alpha. I wonder if that is why?
@@cstevenevenThis is what I thought too
Thanks for making these videos, Jesse! Enormous help. Could you or anyone here explain how K/V= a? I'm a bit confused with that step. :)
Agg physio for 20 years with all the amazing help you have given me wish I could return the favour
This comment is thanks enough! :)
You broke this down really well. Thank you 🙏🏿
Hey Jesse! Thanks for the video! Can you go through the 3D graphical analysis and questions? Especially with state changes such as solid, liquid and gas. Thanks!
Ah yes, that’s a great suggestion! I’ll try to put one together in the next few weeks
Thank You Jesse! Great Teacher
You're welcome
Hi Jesse, thanks for this! What fundamental dimensions would be used for temperature?
Temperature actually is another fundamental dimension. It’s usually represented by K
Thank you so much!
Hi Jesse, i am having a hard time understanding something - why would the dimensions of the second bracket not just be Amps? Seeing as we worked out that Q = amps and the dimension of anything added/subtracted in one expression must all be the same dimension?
Hi Stella, so Q was charge which was equal to zt but Q is not the same as current. This proof is given by the maths and the use of the formula Q = It where I = current in amps
because zt is subtracted from Q, the dimensions of Q and zt must be equal, ie. Q = zt
Using the formula Q = It, we can replace Q with It to give It = zt and then t can cancel out from both sides giving I = z so the dimensions of z are amps.
Hopefully that clears things up!
Hey Jesse, thanks for the video, really helpful. Just at 6:48 I'm struggling to see how you went from M.L.(T^2)/L^2 to M.(L^-1).(T^-2)/L. Wouldn't the L on the top be cancelled out and your just left with L^1 on the bottom. So instead of the final value of P=M.(L^-2).(T^-2), shouldn't it equal P = M.(L^-1).(T^-2)?
Have I misunderstood something?
Hey, I'm not Jesse ahaha but I'll try my best to explain!
So we calculated (M.L.T^-2)/L^2 as the dimensions for Pressure. Just like you said, the L on the top will be cancelled out by the L^2 on the bottom. So a simplified version of this equation will be (M.L^-1.T^-2) which equals Pressure which is what you got too!
Jesse then calculates the dimensions for the alpha (a) constant which has the units of Pressure/Metres. So here we plug in our dimensions for Pressure (M.L^-1.T^-2) and use L as the dimensions for Metres. The final dimensional equation for (a) would then be (M.L^-1.T^-2)/L which is then simplified to (M.L^-2.T^-2) - this is the equation Jesse ends up with at the end. Hope that helps!