Recent content by SimonZ

For hydrogen emission spectra from n = 3 to n = 2: My question is: In the following transition which is (are) possible or impossible and why? 3s --> 2s, 3s --> 2p 3p --> 2s, 3p --> 2p, 3d --> 2s, 3d --> 2p,

B = μo(N/l) * I = μonI is only valid for magnetic field INSIDE the solenoid. The field OUTSIDE the solenoid is zero. So the field is only due to the straight current

Homework Statement A straight wire carrying a current of 42 A lies along the axis of a 6.6 cm-diameter solenoid. The solenoid is 70 cm long and has 250 turns carrying a current of 6.0 A. Estimate the magnitude of the magnetic field 4.2 cm from the wire. Homework Equations magnetic field...

Homework Statement A capacitor and inductor are connected in series to an AC generator. As the generator’s frequency is increased, the current in the capacitor A. increases B: decreases C: does not change D: can't answer this question due to no enough information. Homework...

The question is on www.masteringphysics.com You're right, V = 0, but it won't affect the numerical result (only 2 significant figures needed).

but the website says ir is incorrect! I don't know why.

you have got \theta = \pi/4 - \varphi/2 so 2sin\theta cos(\theta + \varphi) = 2sin(\pi/4 - \varphi/2) sin(\pi/4 - \varphi/2) = 2sin2(\pi/4 - \varphi/2) = [cos(\varphi/2) - sin(\varphi/2)]2 [cos(\varphi/2) - sin(\varphi/2)]2/cos2\varphi = {[cos(\varphi/2) - sin(\varphi/2)]/cos\varphi}2 =...

Homework Statement The flash unit in a camera uses a special circuit to “step up” the V = 3.1 V from the batteries to V' = 300 V, which charges a capacitor The capacitor is then discharged through a flashlamp. The discharge takes t = 12 μs, and the average power dissipated in the flashlamp is...

use gamma r = E/(mc^2) = 1/sqrt(1 - v^2/c^2) so v/c = sqrt[1 - (mc^2/E)^2] for pion, mc^2/E = 0.14/2 = 0.07 v/c = 0.9975

L/2 = r * cos(θ/2)

The Gauss theorem CANNOT be used for finite charged cylinder.

during time interval dt, force F can be regarded as a constant, the work done in dt is dW = F dot ds, power is P = dW/dt = F dot ds/dt = F dot v. so even though force is not constant, P = F dot v is still valid.

yes, but don't put 0.88 in your calculator. Just use calculator for 1/2 arcsin(g*x/vo^2) = 31.2

note that in Fx(t) = k1 + k2y, the y is not a constant, so the integral of k2y dt is not equal to k2yt

q1 = 3.25 nC, q2 = 1.55nC, d = 0.570 m. charge of electron = q, a = 0.100 m, find the final speed of the electron, v = ? initial kinetic energy Eki = 0, potential energy Epi = Kqq1/(d/2) + Kqq2/(d/2) final kinetic energy Ekf = mv^2/2, potential energy Epf = Kqq1/a + Kqq2/(d-a) energy...