Відео

Hello sir ❤️🙏 I have a question to you, why inertia depends upon mass of an object rather than weight (amount of gravitational force act of an object), this is why because mass is a fixed quantity but inertia is not, for example: we put same mass having object at different planets here mass is same but inertia is different i.e due to gravitational force of planets. So sir can we say inertia depends upon weight (amount of gravitational force act of an object)?? Sir if you see so please clear my doubt please❤️🙏

Did we skip198?, it seemed like after 197 came 199th problem

If u die I swear to god I’m going to be the 2nd 🥈 nicest physics teacher😢

10:24 Capacitor

A is square root of r1r2[r1-r2]/2gR^2 here i took GM=gR^2,sorry for that b is 50/3 hrs which is16.67 hrs love the way you teach

Sir, I had a question and I was hoping to get your help. If time(T) , velocity(C) and angular momentum(H) are fundamental quantities then what will be the dimensional formula for mass in context of these?

All concepts Clear, Thank You So Much Legend ❤

T=rs(2(r2-r1)/g) - 33 years

The answers I got were: 2(51R^1.5-6R^1.5)/(3*sqrt(2GM)) 37.04 hours And here is the excel file with the data docs.google.com/spreadsheets/d/1-kjXGWUIBYF2NTrKu74mES6SY-z3yzHgk2DWR5KujFM/edit Hopefully I’m right 😅

Hello, if my answer is true, i can show the complete calculation of a) : a) t=2/3 * (2GM)^(-1/2) *[(R2)^1.5 - (R1)^1.5] b) t=36.28 h Im happy to have you as a friend.

Using the free fall theorem, we can find the time taken from R2 to R1 is a.) t = sqrt(2R²(R2-R1)/GM) And if R2=50R and R1=5R, using the equation above, we can find the exact time is b.) t = 7680 second or t = 2.1 hours I hope it's the correct answer Thanks for the problem, Sir. Can't wait for the explanation 😊

If phi-max for red is 42.4 deg, and for blue it is 40.6 deg, why does the inner part of the rainbow (say at around 41 degrees) actually look blue? Isn't there red light there too because it's inside of 42.4 degrees? Shouldn't it go from red to maybe yellow (as you add green) but once blue gets added that should make the light white? Why is there a visible blue section?

Thanks professor

Thanks professor

Sir Walter Lewin, When you demonstrate the 2100 Hz frequency of sound using the metal rod. you rub the rod with your fingers after you hit it, Why did you do so? 46:37

Answer is time t=√(R2-R1)/g

well what is the reward😐😐☺

Sir I got answer 29.32 seconds for A). Is it correct???

First of all thank you for your lectures ❤ Question - Professor, before reaching at highest point ,thourught the path a=-g because projectile going upside in y direction but after the highest point projectile is going downside in y direction so now acceleration should be a=+g Please tell me?

My answer is approximetly 93,6 hours

Answer for a) 2/3(2GM)^-1/2×[(R1)^3/2 - (R2)^3/2] b) 36.46 hours

Respected walter lewin sir❤ Answer for a) 2/3(2GM)^-1/2×[(R1)^3/2 - (R2)^3/2] b) 36.46 hours Please sir respond to my answers whether they are correct or not! Thank you sir

Sir please start an internship program for high school students so that we could learn more about physics.

Sir , i respect from India up

That one guy hidden on the other side of the room: let's give this thing a good push!

Wow the 200th problem!

Respected professor, how may I send my solution to this problem to you?

To solve this problem, we can use the concept of gravitational potential energy and conservation of mechanical energy. (a) The gravitational potential energy of an object at a distance \( r \) from the center of the Earth is given by: \[ U = -\frac{GMm}{r} \] Where: - \( G \) is the gravitational constant, - \( M \) is the mass of the Earth, - \( m \) is the mass of the object, and - \( r \) is the distance from the center of the Earth. The object falls freely from \( R_2 \) to \( R_1 \), so its total mechanical energy (\( E \)) remains constant: \[ E = K + U \] Where: - \( K \) is the kinetic energy, and - \( U \) is the gravitational potential energy. Since the object starts from rest at \( R_2 \), its initial kinetic energy (\( K_i \)) is zero. Therefore, the total mechanical energy at \( R_2 \) is equal to the gravitational potential energy: \[ E_{R_2} = U_{R_2} \] \[ 0 = -\frac{GMm}{R_2} \] At \( R_1 \), the object has kinetic energy (\( K_f \)) and gravitational potential energy (\( U_{R_1} \)). Therefore, the total mechanical energy at \( R_1 \) is: \[ E_{R_1} = K_f + U_{R_1} \] Using conservation of mechanical energy, we have: \[ E_{R_2} = E_{R_1} \] \[ -\frac{GMm}{R_2} = \frac{1}{2}mv^2 - \frac{GMm}{R_1} \] \[ -\frac{GM}{R_2} = \frac{1}{2}v^2 - \frac{GM}{R_1} \] Where: - \( v \) is the final velocity of the object. Since the object falls freely, its final velocity can be found using the equation for the velocity of an object undergoing constant acceleration: \[ v^2 = u^2 + 2a \Delta s \] Where: - \( u \) is the initial velocity (which is zero), - \( a \) is the acceleration due to gravity, and - \( \Delta s \) is the displacement. The acceleration due to gravity is given by \( a = \frac{GM}{r^2} \), and \( \Delta s = R_2 - R_1 \). Therefore, \[ v^2 = 0 + 2 \left(\frac{GM}{R_2^2} ight) (R_2 - R_1) \] \[ v^2 = 2 \frac{GM}{R_2} - 2 \frac{GM}{R_1} \] Substituting this expression for \( v^2 \) into the previous equation, we can solve for \( t \): \[ -\frac{GM}{R_2} = \frac{1}{2} \left(2 \frac{GM}{R_2} - 2 \frac{GM}{R_1} ight) - \frac{GM}{R_1} \] \[ -\frac{GM}{R_2} = \frac{GM}{R_2} - \frac{GM}{R_1} - \frac{GM}{R_1} \] \[ -\frac{GM}{R_2} = \frac{GM}{R_2} - \frac{2GM}{R_1} \] \[ \frac{GM}{R_2} = \frac{2GM}{R_1} \] \[ R_1 = 2R_2 \] But \( R_1 = 5R_1 \), so \( R_1 = 2R_2 \) is not possible. (b) Given: - \( R_2 = 50 \times R \) - \( R_1 = 5 \times R \) - \( M = 6 \times 10^{24} \) kg - \( R = 6400 \) km (which is \( 6400 \times 10^3 \) m) - \( G = 6.67 \times 10^{-11} \) N m²/kg² We'll use the derived equation for time \( t \) from part a: \[ t = \sqrt{\frac{2(R_2 - R_1)R_1^2}{GM}} \] Substituting the given values: \[ t = \sqrt{\frac{2((50 \times R) - (5 \times R))(5 \times R)^2}{GM}} \] \[ t = \sqrt{\frac{2(45 \times R)(25 \times R^2)}{GM}} \] \[ t = \sqrt{\frac{2250 \times R^3}{GM}} \] Now, let's substitute the values of \( M \), \( G \), and \( R \) and calculate \( t \): \[ t = \sqrt{\frac{2250 \times (6400 \times 10^3)^3}{(6 \times 10^{24}) \times (6.67 \times 10^{-11})}} \] \[ t ≈ \sqrt{\frac{2250 \times (26214400000000000000000)}{4.002 \times 10^{14}}} \] \[ t ≈ \sqrt{\frac{59016000000000000000000000}{4.002 \times 10^{14}}} \] \[ t ≈ \sqrt{14750749.375} \] \[ t ≈ 3840.03 \] Converting seconds to hours: \[ t ≈ \frac{3840.03}{3600} \] \[ t ≈ 1.07 \] Therefore, the time \( t \) it takes for the object to fall from \( R_2 = 50 \times R \) to \( R_1 = 5 \times R \) is approximately 1.07 hours (to 2 decimal places).

Hello sir ❤

Congratulations for the publication of 200th problem. Best wishes for the publication of the 300th one.

Greeting from Paris France. Thanks for your knowledge and enthusiasm in bringing to us. ❤

(a) At a distance 'r' from center, let's say the speed is 'v' We obtain the expression for v by conservation of total energy. (K_i + U_i = K_f + U_f) (Initially, object starts from rest so K_i = 0) U_i = -GMm/R_2, U_f = -GMm/R_1 By further simplifying, we get v = (2GM(1/r - 1/R_2))^(1/2) Also, here v = -dr/dt Now we need to integrate, dr/(1/r - 1/R_2)^(1/2) from R_2 to R_1 on LHS and -(2GM)^1/2 dt from 0 to t on RHS. Put r = R_2*(sin u)^2 and integrate We get R_2(u - 1/2* sin(2u)) = -((2GM/R_2)^1/2)*t Finally, t = (π/2)*( (R_2)^3/2)/(2GM)^1/2 - ((R_2)^3/2)/(2GM)^1/2 * arcsin((R_1/R_2)^1/2) + (R_2)*(R_1/2GM)^1/2 * (1-(R_1/R_2))^1/2 (b) By putting all the given values, time taken from R_2 to R_1 = 0.396 hours.

Oh god after 12th ❤

Find average acceleration (ā) from the two points, 5R and 50R, with a = GM/R^2 Δx = 1/2at^2 Δx = ΔR = 45R => t = sqrt(90R/ā)

Sir i from india , and aspirating for an engineering exam JEE , i belong to general category and in our country their is a huge problem of reservation... This creates trouble for students from general category, thus i request you to post lectures regarding 11th class physics classes i entirely trust ur teachings sir., we just need ur blessings

a) F(r) = m a(r) = m GM/r^2; a(r) = GM/r^2; r(t) = INTEGRAL(a(r) t dt); r^2 dr = GM t dt; r^3(t) = 3/2 GM t^2; then: Δt^2 = 2 (R2-R1)^3 / (3GM); b) R2=50R; R1 = 5R; substituting: Δt = SQRT(2*45^3*R^3 / (3GM)) = 55 hours approx.

Is there a mark scheme for this? I’d love to see the answer

I think the answer is t = 26.48 sec

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I love you sir ❤❤

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Professor , in problem Acceleration due to gravity is considered constant or it is varying?

Love you walter lewin

Hope you are doing well, professor. Greetings from CO!

Yes . I love you very much ❤❤

❤❤first view🎉🎉

Helloo

boring

Neet has biology and chemistry while jee has chemistry and maths. What about them? make videos on them.

its 3am on a sunday night. i love you so much