Recent content by Amar

Yes the equation was incorrectly written and I did the caluclations with the incorrect version. I edited the post and now the equation is Q = Ief2*R*t . I hope this is the expression for the power dissipated you aimed at. However, calculating the result with the correct equation has given me...

Homework Statement An alternating current, with the current value of i = 3*cos314*t, is split into two parts in point A (each part with its own thermal resistance, R1 = 60Ω and R2 = 40Ω), which then connect again in point B. What is the released heat in the part between point A and point B...

Thanks for pointing that out. Yes the 'line of symmetry' should be the normal to the interface ( it's actually called that in my language but I didn't think it would apply to English too). And the angle is closed by the interface and normal. So I'm guessing that my attempt was correct. My...

Homework Statement Moving through air, a wave hits a steady area of water with a angle of 13°. The velocity of the wave in air is 550 m/s and in the water 1650 m/s. What is the refraction angle of the wave? Does the angle move away or to the line of symmetry? Btw. I think I translated it...

Never mind, after another try at the calculation I got that 2.4525 is correct, thanks for the help :)

But what about the double answer I got ? One is 4.905 the other 2.4525 ? Kinda weird that the the other is 2x smaller than the first :P

I must have explained it badly. The height given is 1R EDIT: The title is wrong, sorry..

Homework Statement So the problem I have is calculating the gravity acceleration on the height of 12740km or 2 times the radial distance of Earth. The problem is relatively simple and I think that I have it right but the result doesn't match with the book. ## g_0=9,81 \frac {m}{s^2} ##...

Ahh I see now. I forgot about the 5 and didn't quite understand that concept. Thank you very much :)

After a rather long period of trying different variations I figured it out. It resulted in h = [ sqrt (G x M) / g ] - R This really helped out since it was kinda upside-down... So thank you everyone for leading me to the right solution ( and I'm really grateful for not getting a straight...

Thank you :)

Thank you :) Firstly I calculated the 5x smaller gravity ( g/5 ) which resulted in 1,962 m/s^2 Then I converted the 6370km to meters ( 6 370 000 m ) Then, using my approach sqrt( 1,962 x 6370000 ) / ( 6,67 x 10^-11 x 6x10^24 ) i ended up with sqrt 18383940 / 40,02 x 10^23 which ended up as...

Homework Statement The problem states to calculate the height at which the gravity acceleration is 5x smaller than on the surface of the earth. So g=9,81 m/s^2 R=6370km M=6x10^24 gamma(don't know how to put it :P)=6,67x10^-11 Homework Equations g=gamma x M/(R+h)^2 The Attempt at a Solution...

Hello ! :) My name is Amar, and I am 15 years old. I'm currently in high school and I am really interested in learning and understanding physics. Generally, in my country (Bosnia and Herzegovina) nobody really likes physics (neither did I) but I think that it is essential for my education and...