Me neither 😄

Hopefully it has helped you though!

I'm glad it's helped you! 😊

I'm really glad this video is helping 😊 And don't worry about last weeks test result. I've had a few bad test results in the past even when I thought I'd done really well on the test. Sometimes teachers don't explain all the details that clearly (the same is probably true for my videos hear as well) that's why student's must watch videos from a variety of different channels before it sticks. I wish you luck, Table, for your next test 😁

You're most welcome and thank you for the kind words 😁

Thank you very much 😊

Time is something you can measure and has units of seconds. Dimensional analysis is about making sure that the units on both sides of your equation are the same.

Thank you so much. I'm glad it's helping you 😊

Brilliant! I'm so glad it's helped you out 😊

You're very welcome Jaap. 😊

That's really nice of you to say. Hearing this makes making these videos worth it. 👍

Hi, this is a really good question. There isn't a single, exact method for creating equations in physics and it can be a lengthy process. First we might start by making observations, taking measurements and finding patterns in our experimental data. E.g. We fire alpha particles at a thin sheet of gold and, to our amazement, some alpha particles bounce back! - (Rutherford's Gold Foil Experiment). This result is unexpected, so we use logic and the physics we already know to develop a hypothesis as to why alpha particles fly back in our direction - maybe, gold atoms have a dense "nucleus" where the majority of its mass is located. Even though Rutherford was right - atoms have a dense nucleus, he introduced a new problem and question involving orbiting elections: According to classical mechanics, a charged particle moving in a curved path would emit electromagnetic radiation, causing the electrons to lose energy and spiral into the nucleus. If there is a dense nucleus, why do the electrons still remain in stable orbits in the atom? Rutherford couldn't solve this problem but Niels Bohr attempted with his own experiments and mathematics. Long story short, we need to learn maths and physics sequentially and once our physics and maths knowledge is strong enough, we can develop our own equations and experiments like Rutherford and Bohr.

Thank you. I'm glad it's helped.

@@PhysicsTutoringHub your welcome

Thank you 😊

Brilliant 😊

you should make more videos like this sir

@@oy29 Thanks😊 I've got 79 videos on this channel so far but I'm working on more all the time. I'll be doing more mathematics vids in the near future.

⁠@@PhysicsTutoringHubsir canyou make video on basic of algebraic expression basic

@@PhysicsTutoringHubbtw where ru from ? sir

Hi 😊 That's a good question. So, with this equation, it's best if we break it down into parts rather than trying to understand the whole eqn in one go. r1 and r2 represent lengths. If r1 = 10 ft and r2 = 2 ft, then (r1 - r2) = 10ft - 2ft = 8ft You can see here that our answer for (r1 - r2) = 8ft has dimensions of length (8ft is a measurement along a 1 dimensional line). But, (r1 - r2) has been squared using the ^2 symbol, this means that we are multiplying a length by a length -> (r1 - r2)*(r1 - r2) = 8ft * 8ft = 64ft^2 (64 feet squared) When we multiply two lengths together we get an area. If you look at your bedroom wall, for example, the height of the wall will be around 8ft tall. It's width along the bottom might also be 8ft, which means the total surface area is 8*8 = 64ft^2. This comes in useful if you need to calculate how much paint you'd need to cover your entire wall. Don't worry if you don't understand right away, this video is meant for students aged 16 and above, so you're doing really well getting this far. 😊

Thanks. I'm glad it's helped you.

You're welcome Khadija 😊

I'm really pleased it's helped you Emmanuel 😊

That's realy good. I'm glad it's helped you 😊

Thank you 😊

Hello Nora. I'm really glad the video is helping 😊. I had a quick look back at the part you mentioned. I don't think I said that the sum of r1 and r2 is not a length, but I did say when you multiply two lengths together you'd get an area. So at 18:55 I say that the height itself is a length, but when you square a length, you're effectively multiplying it by itself -- it becomes an area. At 19:35 I say that when we square the sum (r1- r2) it turns into an area, because we are multiplying two lengths together -- (r1-r2)^2 = (r1-r2)*(r1-r2) = L * L = A I hope this clears things up.

Hello. Thank you. Yes, the video is helping a lot. I did not understand much of my lecture but I was able to understand your video. 🙏 Maybe my hearing is wrong, because it still seems to me that from 18:43 you say "...and we can see this, because the sum of r1 and r2 won't result in a length." Anyways, thank you.

@@norawallberg1345 That's okay 😊. I just listened back to this and I did say "will" but my pronunciation of it was a bit off (I don't know why this is, I am a native English speaker 😄). The subtitles do say "won't" which is wrong. So, long story short, I will try and say "will" more clearly next time 😊. I'm really glad the video is helping as well. I always found lectures to be hard to understand. All this physics stuff takes time to fully absorb and understand. So, please keep going with it and don't be afraid to ask more questions in the future and I'll do my best to answer them. Take care.

@@PhysicsTutoringHubI understand! That explains it. Thank you for the inspiring words, I will keep them in mind. Just started studying physics at university. Take care. 😊

@@norawallberg1345 That's brilliant. I wish you the best of luck at uni 😊

Excellent. So glad it's helped 😊

Lol. I'm glad it's helping you. 😊

That's so kind of you to say. Thank you so much 😊

Brilliant, I'm glad it's helped you 😊

When we do order of magnitude calculations, we want to simplify numbers to a power of 10 that closely maches the 67million. So 100 million would be a good choice here. But 50 million is closer to 67 million in this example so I chose 50 million. Looking back at my old physics book, there is an example of estimating the number of breaths a person might take in a lifetime. Now for the the number of days in a year, they have rounded this up to 400 and for the number of hours in a day, they chose 25. These two numbers are much better to use than 100 days and 10 hours respectively.

😁

Excellent. That's usually the way. Take care 👍

Brilliant 😊 So glad it's helping. I'll be posting up Isothermal and adiabatic processes in the next few days.

Thanks so much. That's really kind of you to say. I'm doing the isochoric video next then following up with isothermal and adiabatic. I was going to put them in one video but it got too long so I've split them up. Take care!

Thank you so much 😊

Thank you. I'm really glad it's helped you 😊

Hi Saiful, hope you're well. The answer to your question is yes it would approximate to 5. Let me give you another example where we would round up the order of magnitude in the same way: Let's try and estimate the number of gallons of gasoline used in the USA per year. Let's say we have 1.00x10^8 cars in the USA. Each car travels 1.0000x10^4 mi/yr. Each car can travel 20 miles on 1 gallon (0.05gal/mi) Gallons used per car per year would be: (1.0000x10^4 mi/yr)(5x10^-2 gal/mi) = 500 gal/yr per car Now we can multiply the total number of cars in the US by the estimated gals they use per year: (5x10^2 gal/yr)(1.00x10^8 cars) = 5x10^10 gal ~ 10^11 gal So the order of magnitude is 11. Hope this helps.

A quick way to do it, if the multiplier is less than 3.162 (the square root of 10), the order of magnitude of the number is the power of 10 in the scientific notation, whereas if the multiplier is greater than 3.162, the order of magnitude is one larger than the power of 10 in the scientific notation. Since the multiplier in this case is 5, then the order of magnitude is 4+1=5

You're welcome MD inventions 😊

Thank you 😊. I'll try and slow down a bit. Was there anything in particular that was too fast in this video or was it just the pace overall?

Thank you. I'm realy glad it's helped you 😊

Thank you Meghana 😊

Thank you. I'm glad it's been of help. 😊

Hi, thanks for your kind words. And, brilliant question as well 😊 I’ll do my best to answer your question fully but I’m also planning a video on entropy in the next few weeks. Okay, our hypothetical pendulum is in an isolated space this means that entropy cannot decrease in this scenario (i.e. no outside work can be done to increase the order of the atoms within our system - because, effectively, there is no outside). Entropy will always increase for our pendulum until it reaches a state of maximum entropy - i.e. all the atoms in the system appear to have random positions, velocities, spins etc. Because there are so many atoms/molecules in this system, it’s far more likely to see them in random orientations than in moving in one direction (but not impossible). So, it’s extremely unlikely that the gas molecules will all have the same velocity and push the bob back up to its initial height. If the initial state of our pendulum was a close to 0 Kelvin and the bob was already raised in a gravitational field, the bob has GPE and the system has low entropy. If the bob is released somehow, its stored GPE converts into KE. Molecules of gas, that are very ordered due to their low temperature, will collide with the pendulum and gain KE (the internal energy of the gas increases and so does the disorder - Entropy increases). The same is true for the crystalline structure of the pendulum and pivot. Initially, near 0 Kelvin, the atoms are highly ordered but when gas molecule collide with the structure, they impart a small amount of KE to the solid pendulum and create disorder in the crystalline structure. As this process continues, the whole system will start to increase in temperature and with an increase in temperature there are many more ways for the atoms to orientate themselves. Remember, energy is still conserved here, it’s just been converted into internal energy. And, Once the pendulum reaches maximum entropy (random orientations of atoms), it won’t spontaneously return to its initial state. I hope this helps a bit. I will do a more detailed video/s on entropy in the next couple of weeks including the second law of thermodynamics. Take care and thanks for the interesting question 😊

You're most welcome Sarthak. 😊

Thank you so much Emma 😊

Hi Naren yes you're right, I made a mistake there.The point I was trying to make here, though, was that this dimensional analysis was able to help us come up with one of the equations of kinematics. The constant at this point is an unknown value which we can only obtain through experimental data or geometric reasoning. So, K here is really just a placeholder for an unknown non-dimensional value. Hope this helps.

You're very welcome. I had the exact same question when I was studying physics back in high school. Take care.

Thank you so much 😄.

Thank you so much. Your videos are looking good too 😸. Take care.

@@PhysicsTutoringHub thank you

Cool, ask away and I'll try my best to help.

Hi Bilal. So, yes there is a difference between torque and moments but they use the same equation to measure their value (and have the same units - N.m). With Moments, the force applied to a rigid body at a distance (d) from the falcrum (or the pivot around which a lever turns.) has the "tendency" to rotate the object in question. This word "tendency" is important to note here. It means that it will try to rotate the object/beam but won't be able to. So, think of a diver standing on the end of a diving board. The diver's weight represents a force and that force is applied at a distance (d) away from the diving boards pivot/falcrum. The diving board bends, it doesn't rotate even though it wants to. So this diver's weight multiplied by their distance is the moment. Other examples would be the force applied to the end of a crane's arm. Torque is used in situations when there is an axis that freely rotates. Think of the bolt example at the beginning of the video. Other examples of torque would be the force applied by a bicycle chain on the gears of a bike. I hope this has helped.

I'm really glad it has helped 😁- Brilliant. Take care.