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Would it be because since this takes place in the air, the object isn't experiencing a force from the ground opposite of gravity?

You'll notice gravity is considered initially, but it cancels out, leaving the final answer to no longer depend on gravity.

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.

@@MsProblemSolver thank you!

g=acceleration due to gravity=9.8 m/s^2(approximately)

Glad I could help!

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.

@@MsProblemSolver I think it was asking for a magnitude now that I think about it. Thanks for explaining that magnitude is always positive!

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!

Depends, how fast can you run?

I don't use a camera, just a screen recorder.

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!

Happy to help!

from india, grade(class) 10th

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.

@@MsProblemSolverThanks

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.

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!

That's correct. I create my videos to be easy to follow without added voice explanations.

Great guess. Most of my videos use OneNote but some of my older videos used Whiteboard.

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.

Thank you for your support and Happy New Year!

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.

It knows that isn't less or greater than anyone else in the equation.

I couldn't say. Maybe CZcams thinks you should be a Physicist!

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.

I understand that. You are solving physics, the others will be pretty easy eventually. Keep up the great work. @@MsProblemSolver

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.

That's correct. There is no audio on these videos. While I intend to provide audio in the future, I do not currently have the equipment or skill to provide high quality audio and I believe poor audio is worse than no audio at all. Instead I focus on ensuring my solutions are clear and occasionally add a written comment or directional note when necessary. And I am always willing to answer questions.

That's correct. There is no audio on these videos. While I intend to provide audio in the future, I do not currently have the equipment or skill to provide high quality audio and I believe poor audio is worse than no audio at all. Instead I focus on ensuring my solutions are clear and occasionally add a written comment or directional note when necessary. And I am always willing to answer questions.

I expect he has many years of experience as a teacher. I'll certainly check out his channel.

That's right. My videos currently do not include voice explanations. (While I do intend to eventually provide voice, currently I do not have the equipment or software to provide quality audio and I believe poor audio is worse than no audio at all.) I am careful to ensure the steps are clear without any audio, and include the occasional written note when additional information is needed. I'm also always willing to answer questions.

@@MsProblemSolver thanks a lot man