Beat Physics

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@byronfarrow5995 Күн бұрын

Useful, thank you.

@beatphysics Күн бұрын

Thanks! I'm in the process of releasing more fluid videos too.👍🏽

@therealskmaker 3 күн бұрын

In the first part when calculating the acceleration and taking into consideration friction, I am not sure why you multiplied the force of friction by the weight of the object.

@beatphysics 2 күн бұрын

The formula to find the force of friction is the coefficient of friction multiplied by the normal force. In this case, the weight is equal to the normal force. You don’t take the actual force of friction and multiply it. It’s the coefficient of friction. I hope that helps!

@tinchakxu1584 5 күн бұрын

what if you don't know what angle ball a is traveling from?

@beatphysics 5 күн бұрын

What are you solving for as your final solution? Just the angle? If it's just the angle then you'd have to proceed with the problem solving process, but leave your momentum components in terms of theta and then solve for that at the end. Let me know if this helps and if not give me a little more detail/context of the type of question that you're asking about, thanks!

@RedetzkePhysics 9 күн бұрын

I noticed the slope of the line at position A for KE is positive, but at position C it is zero. I think A should have a zero slope as it should be the same as C. Both indicate zero kinetic energy, but the shapes of the graphs are not the same. The correct shape should be what you have for C. If you slide the graph to the right so C is the starting point, it does not look the same as A currently does. Same is true of A for PE. The slope should be zero at that point.

@beatphysics 8 күн бұрын

Ahhh, you are definitely correct! Thanks for taking the time to comment 👍🏽!

@nosleepdelirium1214 15 күн бұрын

THANK YOU

@beatphysics 15 күн бұрын

You're Welcome!

@connorking1386 15 күн бұрын

I like your content, could you incorporate more visual aids?

@beatphysics 15 күн бұрын

Thanks! Sure, I'll definitely keep that in mind for my future content.

@ipopsiclei6644 16 күн бұрын

What if the rope was a spring and you pushed the system instead of pulling it?

@beatphysics 15 күн бұрын

Hmm good question. I believe if you had a constant force then the spring would take a moment to compress and then maintain that same amount of compression. From then on, I believe the system should be accelerated at the same rate unless that applied force changes.

@kiyoT_T 20 күн бұрын

thank you so much. you saved my whole life!!!

@beatphysics 20 күн бұрын

You're welcome! 👍🏼

@cathrinamarieabella3098 23 күн бұрын

Thank u for saving my life

@beatphysics 22 күн бұрын

haha you’re welcome 👍🏼

@nxmafya1743 28 күн бұрын

why the g isn't -ve?

@beatphysics 28 күн бұрын

The constant "g" is 9.8. You usually include the negative when you are using it in a motion problem as a vector to show the direction of the acceleration, but when it comes to energy, which is not a vector, you would just use 9.8 to develop the magnitude of the gravitational potential energy. I hope that helps!

@bbyeah9720 28 күн бұрын

How would I find Tx

@beatphysics 28 күн бұрын

If you ever have one side of a triangle along with an angle, then you can find any of the other sides. For this problem, once you find Ty (196N), then you can find Tx by doing Tan(30)= Tx / 196N and then multiply both sides by 196N to find your solution. Let me know if you need any more clarification, thanks!

@bbyeah9720 28 күн бұрын

@@beatphysicsdo you have an email? I extremely confused in ap physics and need help knowing if i am doing this right?

@beatphysics 28 күн бұрын

@@bbyeah9720 I don't usually give my e-mail out to people for physics help, but I do reply to all my questions and comments on here. If you need more clarification, then you can continue to ask through this thread. If you feel like you need more help, then I would probably seek out someone to help you in-person. Please feel free to reply with very specific questions about where you're confused and I'll try to be as detailed as possible in my responses, thanks!

@shinichikuroba3999 11 күн бұрын

Tx in this problem would be cancelled out because both T1 and T2 uses cosine. If you try to sum the forces in x-axis, it's T1cos 60 and T2cos 60 and if you try to isolate one tension, it would still end up T1 = T2. That's why you need to solve Ty here to get the tension but you need to do summation of forces in x-axis first.

@danye-c7k 29 күн бұрын

thank you bro

@beatphysics 28 күн бұрын

You're welcome!

@RaffaelloLorenzusSayde 29 күн бұрын

What about applied force? Is there applied force? Also, why did you multiply gravity by two? I thought for the cart gravity force and normal force both cancel out b/c vertically they're opposite forces, and you're only left with the gravity of the hanging object.

@beatphysics 28 күн бұрын

The Fg and Fn acting do cancel, but there is a force of tension applied to the cart, which is similar to an applied force, but not the same because it is applied by a string, so it's considered a force of tension. That force of tension is the same as the Fg experienced by the hanging mass, and since the hanging mass is 2kg then you do 2kg(9.8m/s^2) (mass x acceleration) to solve for that Fg. Let me know if that clears things up or if you have any more questions, thanks!

@jaist Ай бұрын

I forgot how much I enjoyed doing these kind of calculations. Thank you for this video!

@beatphysics Ай бұрын

You’re welcome! 😀

@Humans23 Ай бұрын

I am from India 😅

@nosleepdelirium1214 Ай бұрын

that phrase "increasing in the negative velocity" really cleared it up for me thank you!

@beatphysics Ай бұрын

I'm glad that was helpful!

@navyabirthdayparty7871 Ай бұрын

This was probably the best video I have seen on this topic so far!!!!

@beatphysics Ай бұрын

Thanks a lot, I appreciate the comment! 😀

@CardsNSleevesTCG Ай бұрын

If we solve for the Acute Angle using tan^-1(|y|/|x|) and then calculate the Reference Angle in Standard Notation based on the quadrant that the resultant is in, would we still need to state "N of E" in that situation?

@beatphysics Ай бұрын

Not necessarily, there are many ways to report the correct solution. A lot of times, it may just be the preference of your teacher.

@ManasioMajakMadut Ай бұрын

I have a question

@beatphysics Ай бұрын

Ok, what would you like to ask??

@ManasioMajakMadut Ай бұрын

I have question

@a.meforyou Ай бұрын

how do we know (in the first example) that velocity will be a straight line?

@beatphysics Ай бұрын

All lines on a velocity vs time graph are straight unless you have changing accelerations.

@jamiev9761 Ай бұрын

thank you i have a quiz tomorrow

@beatphysics Ай бұрын

You're welcome. I hope it goes well! 👍🏽

@pl3473 Ай бұрын

Why would you write t the same way you write + 😂

@beatphysics Ай бұрын

hahha you're right! I'll work on that 😀😀

@mayiluvu3 Ай бұрын

At 4:32 why isn't the -18 positive?

@beatphysics Ай бұрын

For that chunk of the formula, it's (2)(-9), so that just leaves a coefficient of -18 in front of the variable.

@mayiluvu3 Ай бұрын

@@beatphysicsokay thanks! I actually passed the exam so this vid helped a lot. thank you so much

@beatphysics Ай бұрын

@@mayiluvu3 Thanks for sharing the good news with me! You're welcome!

@arhea8335 Ай бұрын

Thank you so very much. Have a good day!

@beatphysics Ай бұрын

You’re welcome!

@clashofdudes4814 2 ай бұрын

What if we have uniform speed of 1 car and the otherone is accelerating with some acceleration and no time is given?

@beatphysics Ай бұрын

m.kzbin.info/www/bejne/j57PaKSPq5ZlZ7c

@beatphysics Ай бұрын

I put the link for that type of problem in another reply! It’s a similar process with an acceleration formula substituted in for the constant velocity formula.

@atlasthebastard 2 ай бұрын

can I apply this to hills as well as loops?? like instead of using 2r in mg(2r) can I substitute in the height of a second hill to find the minimum height needed for a rollercoaster to go over the second hill (instead of around a loop)

@EmilySalinasRomero 2 ай бұрын

Thank u !

@beatphysics 2 ай бұрын

You're welcome!

@maticsyt 2 ай бұрын

My goat

@beatphysics 2 ай бұрын

haha thanks!

@venge2761 2 ай бұрын

this helped me out a lot, thanks

@beatphysics 2 ай бұрын

I’m glad it helped! 👍🏼

@adityameena5581 2 ай бұрын

Thanks bro

@beatphysics 2 ай бұрын

You're welcome!

@ParkerDedear-mr9re 2 ай бұрын

thanks for the help bro

@beatphysics 2 ай бұрын

You’re welcome! 👍🏼

@Unroyal-Free 2 ай бұрын

First Please Pin

@beatphysics 2 ай бұрын

Ok you got it! 😎👍🏼

@edloginfaymanuel7855 2 ай бұрын

Quick question where did the 9.8 meter came from?

@beatphysics 2 ай бұрын

That’s the acceleration for any free falling object.

@edloginfaymanuel7855 2 ай бұрын

Sir you are a life saver

@beatphysics 2 ай бұрын

hahah thanks! I'm glad this helped.😀

@dxrkvoyd5696 2 ай бұрын

Is it fixed that clockwise is positive and anticlockwise being negative

@beatphysics 2 ай бұрын

The standard is that clockwise is negative and anticlockwise (counterclockwise) is positive. In the video, I just arbitrarily chose "inside" as positive (clockwise). In a lot of cases the sign isn't that significant as long as you're using opposite signs for opposing torques. Great question!

@dxrkvoyd5696 2 ай бұрын

@@beatphysics Alright thanks!

@Kruisa_ 3 ай бұрын

First! Pin?

@beatphysics 2 ай бұрын

You get the glory of the pin. 😎

@Kruisa_ 2 ай бұрын

@@beatphysics haha ty

@demiyo8424 3 ай бұрын

nice.Subbed

@gamergirlxxx234 3 ай бұрын

what is delta x?

@Fernando-eq3gj 3 ай бұрын

Great video! I have a question about the first pulley (the atwood machine). What would happen if both masses were the same at different heights? The net force should be 0N in both blocks, so no force is moving the system. But the potential energy is different. There is no velocity (both start from rest), so no kinetic energy, but there must be different potential energy between them, so what would happen? Will they move?

@beatphysics 3 ай бұрын

Great question! Right, there's no net force, so the system will be at rest or move at a constant velocity if the system is already in motion. The difference in their initial height isn't really that relevant to the motion they might produce.

@jaist 3 ай бұрын

Amazing video, super clear explanation. What would happen if the "hot air" area would suddenly change its temperature to colder? So on your situation of the palm, having the mirage, then suddently some water falls on the sand, cooling it a bit. What would the observer see?

@beatphysics 3 ай бұрын

Thanks a lot for the comment! Also, great question! The mirage should disappear once the air temperature drops, so it would just depend on how quickly the colder surface would affect the temperature of the air above it, but I think it would fade fairly quickly.

@jaist 3 ай бұрын

@@beatphysics Thanks!

@peachy3278 4 ай бұрын

Question 4 Part a), Isn't the question/the premise of the problem making an incorrrect assumption? The radial distance between Earth and the sphere on the pendulum given in the gravitation formula is (as I know) the radial distance between the center of mass of both objects. If that's the case, given a planet X, which has the exact same mass as Earth just a larger size, shouldn't it impart the same amount of force onto the pendulum as Earth did? The problem can be rewritten as if we take a reference frame directly around Earth (assuming its a perfect sphere), and we take a second reference frame in the shape of a sphere with a slightly larger radius than Earth but still only containing the mass of Earth. Both reference frames "contain" the same mass but only the second is "bigger" as we consider a larger space around Earth. Therefore its obvious to see in this scenario that it would make no sense for the second reference frame to have any done any more or less work than Earth on the pendulum.

@beatphysics 4 ай бұрын

Hi! If im understanding your question correctly then I believe you’re not considering the larger planet has a larger radius, thus decreasing the force of gravity on the pendulum.

@peachy3278 4 ай бұрын

@@beatphysics Hi! thanks for responding :) why exactly would the force of gravity on the pendulum due to the larger planet with the larger radius be less than that due to Earth? I thought that the force of gravity between two masses is proportional to the 1/r^2 where r is the radial distance between the center of mass of both masses. Thanks!

@beatphysics 4 ай бұрын

@@peachy3278 everything you're saying is exactly right, but if the new planet is larger, then its center of mass is going to be farther away due to its larger radius. If you were to pretend Jupiter had the same mass as Earth, then the force of gravity would be significantly less because the center of mass is much farther from the surface.

@peachy3278 4 ай бұрын

@beatphysics Ohh right sorry, for some reason i didnt understand that the pendulum would of course be on the surface of both planets (i have no idea what else it couldve been) and believed that the pendulum would be placed at the same distance away from the com in both setups. thanks!

@Abiy-yasb 4 ай бұрын

Loved it!!

@beatphysics 4 ай бұрын

👍🏼 Thanks!

@fornx8574 4 ай бұрын

no solid explanation of the formulas build and can't identify the operators between terms. This was challenging to watch

@beatphysics 4 ай бұрын

Thanks for the comment. Let me know if you want me to clarify an idea for you!

@haloitsmeh9120 5 ай бұрын

Im pretty sure i failed this exam :l

@haloitsmeh9120 5 ай бұрын

So, I had put the Fh force going into the bar itself, does it still count for points or would I lose points for not matching it to Ft? Technically, it is going in the opposite direction of FT but it's going inside the bar because I misinterpreted what I read for the scenario.

@beatphysics 5 ай бұрын

It depends on how they grade it, but that seems like a minor portion of the problem. So i would guess it’s either right or wrong with no partial correctness considered.

@haloitsmeh9120 4 ай бұрын

Thank you!

@plankstone 4 ай бұрын

So what’d you get?

@KavonThompson 5 ай бұрын

That intro is crazyyyyyyy❤‍🔥

@beatphysics 5 ай бұрын

@kavonthompson lol No one has ever commented on the intro, thanks🙏🏽!

@Edgingtutor 5 ай бұрын

Oh my God. On the experimental part I read over “use only in the instruments given” i’m screwed

@Edgingtutor 5 ай бұрын

Would I get partial credit because I was correct or just a complete 0? 😭😭

@beatphysics 5 ай бұрын

I think it will be super common that people used Hooke’s Law and a meter stick. I think you may possibly get a point for correctly identifying how to find the spring constant with different instruments and for reducing uncertainty in the lab. Other than that, I’m not quite sure how they’re going to award points for that section.

@genekachar5747 5 ай бұрын

Does anyone know if the AP Physics Late testing exam FRQ gets released?

@beatphysics 5 ай бұрын

Good question, not that I’m aware of.

@1974shikha 5 ай бұрын

These questions were not in ap physics 1 2024

@beatphysics 5 ай бұрын

Not all the versions are identical, but most of them are the same.

@jonathanma2347 5 ай бұрын

thanks, I feel confident for BC calc now!

@beatphysics 5 ай бұрын

Awesome 😎

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