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I agree. Please don't watch this. 😉

The area under the curve tells us the displacement/location of the object. When the 2 areas are the same then the objects must be at the same location. For example (pause the video at 13:22), the location of the blue line would be 22.5m from its starting point at 3 seconds (the area of the red triangle.) If the green object was traveling at a constant velocity of 7.5 m/s, it would also be at a location of 22.5 m in 3 seconds (7.5 m/s x 3 s = 22.5m) Hope that helps.

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erm what the sigma

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on skibidi no cap frfr ong

Thanks. I appreciate that!

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on skibidi

Thank you!

@@pghsphysics i gotchu C:

You're welcome! Glad you found it interesting.

Thank you, Emu! I appreciate it.

Thanks for the kind words. Glad I could help. Good luck!

You are correct that speed is just magnitude (a number) and can't be negative. However, velocity is a vector that gives both the speed and the direction of travel. In this problem, we are defining the upward direction as positive and the downward as negative. Therefore, when I throw the ball up it has a velocity of +5m/s and right before I catch it, it has a velocity of -5 m/s. You can also specifically state the direction instead of using +/- (For example, 5 m/s downwards) Velocity and displacement are different quantities and so do not need to have the same magnitude.

No problem. Glad to help.

Thanks for noticing! You are correct, should be 2 ⅔ .

Yes, assuming that the machine is still pulling at 30 degrees to the original horizontal, then once the box goes on the plane the new angle would be 20 degrees and you would use (100N)(10m)cos(20) to find the work done by the machine. To find the net work, you would subtract the work done by the component of gravity down the plane (and friction if it is present) Hope that helps.

Always nice to hear from the rice president from time to time.

Not exactly, but I do have one that finds the force. To find the work, simply multiply by the distance traveled. ruclips.net/video/ipJVYnLMWVo/видео.html

Hope it went well!

Glad it helped!

Don't remember. Either Cutnel, Serway or Giancoli.

No problem. Good Luck!

Very nice. First person in 5 months to notice. 😀

You got it!

Glad it helped!

It is Newton's Gravitational Constant = 6.67E-11 and is apart of the gravitational equation. It was found experimentally.

I am going to try to make one per week leading up to the exam.

Good question. Work is equal to the change in energy. In this problem, the object is moving at a constant velocity, so that the KE is not changing. All the work is turning into gravitational PE and thermal energy from friction. Technically, there must be a bit of extra work to get the object moving in the first place. Though, this would be cancelled out by the negative work when it stops at the top. Practically though, the crux of the problem occurs during the constant velocity phase and KE is ignored in the problem.

Hope you did well!

Glad it helped!

Glad it helped!

Will there be a different formula for that

You calculate it similarly but use the vertical component of gravity, rather than the entire velocity. ie. Vy =28sin30 = 14 m/s then use the Vf2 = Vo2 + 2ad to find max height using 14m/s as Vo Here's a video I made about this : ruclips.net/video/T026vMJHo3k/видео.html

Should be fine.

No, M3 doesn't contribute to the force in any way. It just adds inertia to the system which would decrease the acceleration.

Hi. I believe that you made a very common calculator mistake (in fact, I addressed this with my class just last week.) If you type this into your calculator --> 25/2(9.8) then your calculator will first divide 25 by 2 and then MULTIPLY by 9.8. Instead, you should make sure that you put the denominator in parenthesis --> 25/(2x9.8). You could also divide twice --> 25/2/9.8. Or you could first multiple the denominator 2x9.8 = 19.6 and then divide 25/19.6=1.28. Hope that helps. Its good that you made this error early in your career. It will save you all kinds of time in the future.

Hi Cecilia! Glad to hear that you are doing so well. Good Luck!

That's a good question. A centrifugal force is actually not a force at all, its just a "feeling" of force. There is nothing actually pushing you against the wall. In fact, it is the wall that pushes against you and keeps you moving in a circle which we call centripetal force (imagine what would happen if the wall suddenly disappeared...you would no longer move in a circle.) The faster you move, the harder the wall pushes on you, the harder you push back on the wall, and this "feeling" of force increases. Here's another example... you are driving in a car and then suddenly turn to the right. It would "feel" like you are being pushed to the left...but nothing is pushing you left. You are actually trying to continue moving in a straight line (Newton's first law) and the door (or seatbelt) keeps you from falling out of the car (it pushes you right.)

The KE we were looking for is the bullet-block combination (after the bullet was imbedded.) The reason you can't use the KE of the bullet before it collides with the block is that there will be a tremendous amount of KE lost due to the collision, the amount of which is not able to be determined.

In this problem the applied force is parallel to the surface, so there is no component into the surface. However, if you pulled up at an angle to the surface, then the Fn would be mgcos(theta) - Fsin(angle). Here is another video that covers this concept. ruclips.net/video/ipJVYnLMWVo/видео.html

Here's a video you might like. ruclips.net/video/JYqfVE-fykk/видео.html