Currently focusing on producing clearly written problem walk-through videos, which do not include or depend on the addition of voice explanations. Why no sound? Two main reasons. First, adding sound will slow me down immensely and I have a huge list of subjects I want to work through. This is a passion project and my time is limited. Second, accessibility is very important to me. By not relying on voice explanations, my videos are more accessible to various audiences for whom spoken English is not a primary language.
I have complete the high school AP level Physics problems (algebra-based) minus Atomic and Subatomic Particle topics, which I may add in later. Topics towards the end of the curriculum are still trickling in as I review the videos for accuracy.
I am currently working on: College Physics (Calculus Based) and Calculus.
Up next: Classical Mechanics, Algebra, Linear Algebra
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While I do have plans to add audio, I'm not there yet. I don't have the equipment or experience to add high quality audio just yet. Until then, I will continue to do my best to ensure that my walk thoughts are clear even without verbal explanations.
F is 104N, which is the magnitude of the force. Fx is given in the problem as 90 N, which is the x component of the force (the component parallel to the ramp).
I get questions from a number of different books and sources, and you will find some of the same problems in different places. That being said, I 'think' this one is from Young/Freeman University Physics.
I try to make my solutions clear without the need for verbal explanations. While I may add these in the future, I am currently focused on expanding my library of solutions.
Both part b and part c follow the exact same steps and equations as in part a. First convert the frequency to Hz (1 MHz=10^6 Hz) and then input into the equation.
This problem starts with some initial forward velocity when it arrives at the hill (Like someone gave it a good shove forward and let go, so there is no ongoing force in that direction.) . The hill changes the direction of that velocity, but the magnitude remains the same, which starts it up the hill. But now it has the force of gravity in the downward direction and the force of friction in the direction opposite motion. Eventually with no other forces, the block slows down, stops and starts moving in the opposite direction.
I believe this one is from Young&Freeman-University Physics. I have used several sources for problems, and you will often find the same or similar problems in many places.
You're right! Good catch. Based on the numbers, this should be negative. It is 230 vs 230.4 due to number of significant figures, but I would still expect to see the negative. I'll have to take a look at the original problem to see if there was some additional context I failed to include. Sometimes they are looking for a magnitude, which would be a positive number with a description of direction. It's also certainly possible I made a mistake. I'll review this problem in the next couple days and either make a note to explain the difference or remove it from my channel. Great job catching this one! And thanks for pointing it out; I do my best to correct mistakes when reviewing and editing, but occasionally I miss something.
I do have plans to add voice in the future. In the meantime, I am doing my best to ensure my walk-throughs are clear even without added audio. Thanks for your feedback!
cos(45)tan(30)+sin(45)=0.115; 1/0.115=0.897; w/(cos(45)tan(30)+sin(45))=0.897w; Some of the most common issues are simple calculator issues and I have made TONS of them over the years. Good job double checking. Hope this helped.
I do have plans to do this in the future. In the meantime, I am doing my best to ensure my walkthroughs are clear even without added audio. Thanks for your feedback!
Average acceleration is a constant acceleration. If you look at the time from t=10.0s to t=20.0s as a whole, you can use the constant acceleration equations. You can only look at the period as a whole, since the instantaneous acceleration and velocity may deviate from the average. But the constant acceleration equations can apply . There are almost always several ways your can derive a solution; you may have used a different method and came up with the same answer.
That's correct. I try to make my walk-through videos understandable without verbal and minimal written explanations. The math should speak for itself. I am looking into adding sound at some point in the future.
Hello! Your videos are really helpful to me. I'm glad you decided to make this channel for those students who are studying/practicing physics. But I hope you can use your voice and explain so we can understand better. Anyway, thank you!
I'm glad my videos are helpful. I do intend to add voice explanations at some point, but high quality sound requires some equipment and software that I am still researching. Right now, I am focusing on creating clear walkthroughs that can be understood without the added voice explanations. I'm also always willing to answer questions. Good luck with your studies!
The selection of which equation to use is based on which variables are available and what you are looking for. This could have been solved in a few different ways. This is just the path I chose. As long as you apply the equations correctly, any path should take you to the same result.
For this problem, vertical displacement d=vt+at^2/2. Since a=-g, and initial velocity is zero, we use d=at^2/2. For horizonal displacement where a=0, we would use d=vt.
The fantastic part about physics is that even if you take another path, as long as all of your assumption are correct, you will come to the same answer. I took a different path in my video and came to the same answer as you. Great job!
That's correct. My videos do not currently include voice explanations. I do my best to ensure my walkthroughs are clear without additional verbal explanation as I do not currently have the software, equipment or experience to provide high quality audio.
I appreciate the support. I do plan on eventually adding audio, however I am still working on figuring out the appropriate software and hardware necessary to provide high quality audio. In the meantime, I am focusing on providing clear walk-throughs that can be understood without any audio.
Pls can you help me out on this three charges of 2 nc, −5 nc, and 0.2 nc are situated at p(2, π/2, π/4), q(1, π, π/2), and s(5, π/3, 2π/3), respectively. find the force acting on the 2-nc charge at point p. is this a force of attraction or repulsion?
Unfortunately I haven't gotten to the more complex problems yet , so I don't have a walk-through to point you to. This is going to be solved in a very similar manner to the problem shown. Find the Electric force of both charges with the charge at point p. Then sum the components and use components to get final magnitude and direction. The trick here is going to be that this appears to be in 3-D polar coordinates. You will want to project this onto the x-y-z coordinate system and then you will be solving the problem in x-y-z instead of just x-y like the given problem.
Yes, but not immediately. Maybe in a couple months. I want to spend some time exploring some different software and testing everything out. I also don't have a lot of time in a quiet space right now, so that would severely limit how many videos I could produce. But they ARE coming.
