Recent content by TJDF

some progress... so e = 1.60e-19 C f will equal 0... so... Vo = θ/e ; θ = e*Vo = (1.60e-19J)*( ? V) hmmmm... how to figure out Vo?

Hmmm... still no luck on this one yet. ideas?

Homework Statement The American physicist Robert A. Millikan (1868-1953) won the 1923 Nobel Prize in physics, in part for his work on the photoelectric effect. Assume that Millikan observed for a hypothetical metal a maximum kinetic energy of 0.535 eV when electrons were ejected with 431.7 nm...

oops, forgot the 5%, got it now.

oops... 3.3297e-28 is actually 3.3297e-19... but that gives me 2.252e20 photons/s, which is still marked as incorrect, am I still missing something?

Homework Statement I have another question, An incandescent light bulb consuming 75 W emits only 5% of this energy as visible light of wavelength 597 nm; the frequency of the emitted light is 5.02e+14 Hz. How many photons per second of this light does the bulb emit? Homework Equations...

I figured out c too! Thank you so much for your help!

Okay, I figured out b... I thought, since a J/s is a W... Then the energy, (6.341825e-3 J)/0.000456s = 13.9076 W... so far, so good... now, c.

Ah, that's very interesting! I never though of using that 100°C boiling point. so, with that... Energy to Vaporize = Q = m*l = (2.5e-9 kg)(2.256×106 J/kg) = 5.64e-03 J and Q = m*c*delta(t) = (2.5e-9 kg)(4190 J/(kg·K))(67K) = 7.01825e-4 J adding these two values... Q1 +Q2 = 6.341825e-3 J, which...

Okay... I'm starting to make some progress, but I keep getting stuck in my ideas... Energy to Vaporize = Q = m*l = (2.5e-9 kg)(2.256×106 J/kg) = 5.64e-03 J... so... Q = m*c*delta(t) ; delta(t) = Q/(m*c) = 5.64e-03 J/[(2.5e-9 kg)x(4190 J/(kg·K))] delta(t) = 538.42 K but... I don't understand...

Homework Statement A pulsed dye laser emits light of wavelength 576 nm in pulses of 456 µs duration. This light is absorbed by the hemoglobin in the blood and can therefore be used to remove vascular lesions, such as certain blemishes and birthmarks. To get an estimate for the power required...

Thanks so much! I understand it perfectly now!

Wait... are the correct answers C, D, and E? C is true because q1 is a negative charge. D is true because the contour lines are closer together, and D is closer to the positive charge meaning the field must be stronger because... the charge is stronger. E is true because q3 is the only positive...

Is the electric field zero if the equipotential contour line is 0 volts? How does the diameter of the circle affect the strength of the charge? Are electric fields on the same contour line equal or judging by the fact that closer equipotential lines are, then stronger gradients, then can they be...

Thanks. I still don't know how each would apply to the questions though. A) The electric field at g is zero. ---> I can't tell whether this is true or not, I'm starting to have my doubts, but I can't explain why. B) Charge Q2 is the largest negative charge. ---> I was told the...