Recent content by hellowmad

thank for checking

Yes it is want I mean. Thanks.

The angular position of t)he first diffraction minimum is θ≈sinθ= λ/a, and dsinθ=mλ, so m = (dsinθ) /=[d(λ/a)]/λ =d/a = (2.4 x 10-4 m)/(8.0 x 10-5 m) =3. Since both bright and dark pots separated on both sides of central bright region, so the smaller bright spots observable within the central...

Thank you so much kuruman

Yes, this is the point I post here to see if I forget something or make a mistake.

Thank TSny. Yes couple typo. Sorry for that. My answer is 4.05x10^6 V/m. Thank you everyone

No matter what charge places on the center of triangle, the E field direction of two -4µC's Y-direction is the same as that of 2µC one. The final E-field is the sum of them, So I take the absolute values of the charges here. Thanks

It is not modulus sign. it just means absolute number only. So, it stands as 2µC and 4µC.

I've found the distance from each point to the center, which is equal to r=20x1.732/3 = 11.55 cm. I find out that E2 and E3 due to -4µEyC on x-direction canceled each other. The E2y = E3Y = EY = E2Ycos60 = E2/2 = [(KQ2)/r^2]/2 So the net E-field: E = E1 +E2y+E3Y =kQ1/r^2 + [(KQ2)/r^2]/2 +...

got it. I forgot it. in that case the different is zero as both share the same initial and final conditions. Thank you! great help!:smile:

Question: Two samples of a monatomic ideal gas are in separate containers at the same conditions of pressure, volume, and temperature (V = 1.00 L and P = 1.00 atm). Both samples undergo changes in conditions and finish with V = 2.00 L and P = 2.00 atm. However, in the first sample, the volume is...