PepperEng Tutor
PepperEng Tutor
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Відео

L'Hopitals Rule with Composite Exponential Functions (The Limit Chain Rule)
Переглядів 7 тис.4 роки тому
How to change indeterminate exponential forms into L'Hopitals indeterminate forms and evaluate the function.
Compounds Dielectrics - Series or Parallel?
Переглядів 1,1 тис.4 роки тому
0:00 - Theory 1:25 - Example 1 9:13 - Example 2 18:00 - Example 3 Example 1 As shown, three square metal plates A, B, and C each 12.0 on a side and 1.5 mm thick. The plates are separated by sheets of paper 0.45 mm thick and with dielectric constant 4.2. The outer plates are connected together and connected to point b. The inner plate is connected to point a. a) Show the charge distribution on t...
Right Hand Rule - Explained!
Переглядів 5334 роки тому
0:00 Current and field convention 1:14 'Wire rule' 2:16 'Solenoid rule' 3:23 'Particle rule' 5:27 Summary of Right Hand Rules 6:05 Pictures of Right Hand Rules 8:49 Question 1 - Spiralling particle 12:00 Question 2 - Magnetic field of a rod 14:22 Question 3 - Loop of wire 16:10 Question 4 - Rod on a table Question 1 - Spiralling particle A uniform magnetic field is directed into the page. A cha...
Basquin's and Coffin's Laws, Fatigue
Переглядів 12 тис.4 роки тому
Basquin's and Coffin's Laws, Fatigue The high-cycle fatigue data is as follows: Using the data above, estimate the maximum stress amplitude to ensure 10^8 cycles.
Statics Bell Crank | The bell crank system shown in the figure below is attached to a horizontal...
Переглядів 2,3 тис.4 роки тому
The bell crank system shown in the figure below is attached to a horizontal spring BC with stiffness k = 2 kN/m and a pin support at point A. Spring BC is unstretched when angle 𝜃 = 0°, and the bell crank is in equilibrium under the effect of the external force F when angle 𝜃 = 15°. If spring BC remains in a horizontal position at all times due to the roller at C, determine the following when t...
Everything About the Unit Circle (6 minutes!)
Переглядів 684 роки тому
Everything you need to know about the unit circle, but in 5 minutes! Special triangles - www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-special-right-triangles/v/30-60-90-triangle-side-ratios-proof 0:00 - Introduction to the Unit Circle 0:48 - Introduction to Coordinates (Special Triangles) 1:50 - Coordinates 2:33 - Introduction to Radians 2:54 - Radians 4:13 - Example Problem
Absolute Value Functions as Piecewises
Переглядів 774 роки тому
Converting absolute value functions into a piecewise 0:00 Absolute values introduction 1:15 Piecewise introduction 4:30 Example 1 6:35 Example 2 9:45 Example 3 11:42 Example 4
Flowed text not working! (Inkscape, justifying text)
Переглядів 12 тис.4 роки тому
How to block-justify text in inkscape.
Hardest Statics Question | If F = {100k} N, determine the couple moment that acts on the assembly...
Переглядів 1,6 тис.4 роки тому
If F = {100k} N, determine the couple moment that acts on the assembly. Express the result as a cartesian vector. Member BA lies in the xy-plane. From Chegg.
Integrating Electric Field for a Semicircle | A semicircle of radius "a" is in the first and...
Переглядів 20 тис.4 роки тому
A semicircle of radius "a" is in the first and second quadrants, with the centre of curvature at the origin. Positive charge Q is distributed around the left half of the semicircle and negative charge -Q is distributed uniformly around the right half of the semicircle. What is the magnitude of the net electric field at the origin produced by this distribution of charge? DISCLAIMER: Question tak...
Integrating Electric Field for a Sphere | A solid sphere of radius R contains a total charge Q...
Переглядів 2,6 тис.4 роки тому
A solid sphere of radius R contains a total charge Q distributed uniformly throughout its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the “self-energy” of the charge distribution. DISCLAIMER: Question taken from University Physics 14th Edition
Integrating Electric Field for a Charged Rod | A positive charge Q is distributed uniformly along...
Переглядів 17 тис.4 роки тому
Positive charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positive x-axis at x = a r, a distance r to the right end of Q. (a) Calculate the x- and y-components of the electric field produced by the charge distribution Q at points on the positive x-axis where x is greater than a. (b) Calculate the force (magnitude and direction) that...
Understanding Electric Field and Potential Integration
Переглядів 7944 роки тому
Why do we integrate in physics? How to integrate electric fields and electric potential in physics? 0:00 Electric field 4:11 Electric potential 6:13 Steps

КОМЕНТАРІ

  • @spertica
    @spertica Місяць тому

    Grazie, funziona! Non capisco sinceramente tutta questa involuzione per fare una cosa banale, pollice meno per Inkscape su questo aspetto!

  • @KrishayIyer
    @KrishayIyer 4 місяці тому

    Hi, I'm doing this problem from a textbook. For Part A, the textbook says the correct answer is kQ/x(x-a); there is apparently no inclusion of r. Does this have something to do with the problem stating x>a?

  • @chem-fun7866
    @chem-fun7866 6 місяців тому

    Sir I am from India In my inksacape 1.2 not working your trick

  • @TriThom50
    @TriThom50 7 місяців тому

    Shouldn't the integral be from a to a+r

  • @georgeogbuma1472
    @georgeogbuma1472 8 місяців тому

    Please continue your channel the world needs you

  • @MCoolguy
    @MCoolguy 11 місяців тому

    love you man

  • @lecctron
    @lecctron 11 місяців тому

    you saved my life

  • @khansaboo2030
    @khansaboo2030 Рік тому

    how tf it became -1/x??

    • @FloppaEnjoy
      @FloppaEnjoy 11 місяців тому

      that is the integral of 1/x^2

  • @JesseD-hq6ne
    @JesseD-hq6ne Рік тому

    For the charge at the end, if it were the same charge everywhere, would lambda = 2Q/pi*r ? Also, where did you get the 2Q on top if there is no 0.5*pi*r below the Q?

  • @wandersondebrites5977
    @wandersondebrites5977 Рік тому

    Thank you

  • @robertosarquis5219
    @robertosarquis5219 Рік тому

    Thank you! 😀

  • @gilr.7
    @gilr.7 Рік тому

    Thanks a lot for this video!

  • @piey1737
    @piey1737 Рік тому

    2:00 Sir, I have question 1. I want know How to calculate Right side equation Which paper is it from?

  • @saniyakaale6727
    @saniyakaale6727 Рік тому

    Could you please solve this equation without using mathematica 0.005= (490/159195)*(2Nf)^-0.065 + 0.463*(2Nf)^-0.65

  • @hadhad129
    @hadhad129 Рік тому

    Where is the 2 from I keep getting 2 not 4 at the end

  • @QuantitativeDiaries
    @QuantitativeDiaries Рік тому

    Great explanation! Love the granularity

  • @HussainAli-hn2ef
    @HussainAli-hn2ef Рік тому

    amazing, thank you

  • @NMKM25516
    @NMKM25516 2 роки тому

    At 0:30 why did you take log(192/167) - b? shouldn't it be log(192/167) +b..

  • @sunnydays6144
    @sunnydays6144 2 роки тому

    thank you so much, you saved my life in engg calc this week

  • @zDoubleE23
    @zDoubleE23 2 роки тому

    I don't see how this works with your cos(theta) trick because that is only for right triangles, not arc lengths. If you used cos(theta) with that radius, you cut yourself short on the x axis.

  • @Matoro342
    @Matoro342 2 роки тому

    My man. Thank you.

  • @fahimshahriar6576
    @fahimshahriar6576 2 роки тому

    Thanks

  • @mithilarajeshirke4193
    @mithilarajeshirke4193 2 роки тому

    Do you have any references for this particular formula? As I am getting the formula in various journal articles in a different way such as - S = A (Nf)^b, so here direct S is taken, while that of in your video, delta S (Smax-Smin) is taken.

  • @rumanaislam4758
    @rumanaislam4758 2 роки тому

    This video was really helpful 👌

  • @Danielagostinho21
    @Danielagostinho21 2 роки тому

    i think the greater than and smaller than signs are switched. Shouldn't High cycles be when the cycles is greater(>) than 10^4 and low cycles smaller(<) than?

  • @danielnidan
    @danielnidan 2 роки тому

    This solution is valid... and I believe it is a good approximation. However, the extension of the spring is not exactly as given due to the fact that point B travels slightly vertically (y direction) as well as horizontally (x direction). Thus the extension is slightly more... and the resulting force on the spring will be slightly more.

  • @DregerProductions
    @DregerProductions 2 роки тому

    What a unit

  • @ayaadel3250
    @ayaadel3250 2 роки тому

    thank you so much this video is extremely helpful , but I was wondering why we treated the electric field as it is from one source ,, like I thought the the X-component of each electric field (one from -Q and +Q) is going to cancel each other and we are going to only integerate to get the y component ( Sin will be in our integeration instead of cos) ,,so why we did only use COS like there is no other component. thanks in advance.

  • @angelasuaza9512
    @angelasuaza9512 2 роки тому

    Thank you so much! I suscribed your channel :).

  • @wren4077
    @wren4077 2 роки тому

    why is stress range twice the amplitude

  • @stella-hg7gl
    @stella-hg7gl 2 роки тому

    Thank you so much for this video!! I really needed it and you're very good at explaining

  • @jacklavorato7111
    @jacklavorato7111 2 роки тому

    Lets get into it #peppereng

  • @jacklavorato7111
    @jacklavorato7111 2 роки тому

    nice

  • @jacklavorato7111
    @jacklavorato7111 2 роки тому

    Crazy twist at the end

  • @jacklavorato7111
    @jacklavorato7111 2 роки тому

    not normally but often great content

  • @jacklavorato7111
    @jacklavorato7111 2 роки тому

    This was so helpful I passed my class because of you and your voice sounds so hot

  • @ayaosigodfrey7591
    @ayaosigodfrey7591 3 роки тому

    I am using inkscape version 1.1 and its not working

  • @badgalirri
    @badgalirri 3 роки тому

    At 20:00 how do you get d/dx (x^3sec^2(x)) to be like that? You say you apply the chain rule but shouldn't you apply the product rule? d/dx(f(x)*g(x)) = d/dx(f(x)) * g(x) + d/dx(g(x)) * f(x). I mean d/dx(x^3sec^2(x)) is 3x^2sec^2(x)+2sec^2xtanxx^3. The bottom derivative is also not correct. It is not 2sec^2(x) but rather -2sec^2(x). Seemingly great video up until that point though.

  • @usa6486
    @usa6486 3 роки тому

    fuckingHard!!! Thank you very much.. :)

  • @nicholasrodriguez1234
    @nicholasrodriguez1234 3 роки тому

    Thanks! Good work

  • @AJAXIMOS
    @AJAXIMOS 3 роки тому

    unfortunately, it didn't work for me :( any different ways to correct this ? I tried to draw the text box before typing, the symbol of block-justify turned to be on, but still, when i click it, it doesn't block-justify the text :/

  • @alemercado1751
    @alemercado1751 3 роки тому

    The angle calculations for alpha and gamma must be zero because the angle between the x and y axis is 90

    • @pepperengtutor8529
      @pepperengtutor8529 3 роки тому

      Ah yesthat's an error in my diagram. Thanks for pointing that out! The 60deg should be from the purple x-axis to the blue bar. With this correction, the alpha and gamma angles are not 0.

  • @royale9985
    @royale9985 3 роки тому

    Thank you so much for this!!

  • @beckettmurtaugh3170
    @beckettmurtaugh3170 3 роки тому

    why is he pronouncing ln like "lawn"

    • @Jason-ih3ee
      @Jason-ih3ee 6 місяців тому

      thats how its supposed to be pronounced

  • @mileshunter5285
    @mileshunter5285 3 роки тому

    2.8 thousand views, damn

  • @peterfisher3161
    @peterfisher3161 3 роки тому

    Worked, but only when I inserted the text in a box, as you did, just a text object did not get this option.

  • @mili75005
    @mili75005 3 роки тому

    Thank You!! super helpful :)

  • @rubyash1036
    @rubyash1036 3 роки тому

    why this way doesn't work in my inkscape

  • @amyschmitz2138
    @amyschmitz2138 3 роки тому

    What version on Inkscape are you using? I can't edit my text preferences in Inkscape 0.92

  • @konnerbrickey9569
    @konnerbrickey9569 4 роки тому

    Thanks!