Recent content by Qwurty2.0

Thanks, I was looking at the circuit wrong (I thought the charge was coming out the top and hitting the resistor after the capacitor). Is there a way to calculate the voltage of the capacitor as it is discharging from 5V?

Homework Statement Calculate the voltage, current, and power across the two resistors for every second of run 1 in the RC circuit. The RC Circuit: The settings of the circuit: The Graph. The capacitor is charged until it reaches 5.0 V and then discharged until it reaches 1.0 V. This...

I'll fix my units. The answer for question (a) is correct according to the book solutions. EDIT: I misread the answer, it is 0.2 ps. So there is no problem (besides my units).

Homework Statement Two positive charges +Q are affixed rigidly to the x-axis, one at x = +d and the other at x = -d. A third charge +q of mass m, which is constrained to move only along the x-axis, is displaced from the origin by a small distance s << d and then released from rest. (a) Show...

Ah, my mistake. I understand what I was doing wrong. Thank you.

But if 20°C (293.15 K) is the initial temperature and the final temperature is Ti + ΔT = 20°C + 25°C = 293.15 K + 298.15 K = 591.3 K, and ΔT is Tf - Ti, then isn't ΔT = 591.3 K - 293.15 K = 298.15 K? What am I missing? Don't they give the change in temp right away? :(

I calculate that ΔSSi = -72.939 J/K and ΔSHg = 77.697 J/K. Added them together, I got a net entropy change of 4.758 J/K.

ΔS = ΔSH + ΔSL = -Q/THM + Q/TLM ΔSSi = - (85394.4 J)/(321.96 K) = -265.23 J/K ΔSHg = (33153.12 J)/(321.96 K) = 102.97 J/K ΔS = -265.23 J/K + 102.97 J/K = -162.26 Am I supposed to use the middle heat (321.96 K) to calculate the ΔQ for each substance (ΔTSi = 333.15 K - 321.96 K = 11.19 K and...

I believe so. There is 1 significant figure (the 1 liter volume), so the answer I would have is 2x10-4. But the issue isn't the sig. figs. (at least I don't think), it's the exponent. My answer has 2x10-4, while the closest answer is 2x10-5. I've gone over my calculations and I believe I have...

Homework Statement The coefficient of volume expansion of olive oil is 0.68 × 10-3 K-1. A 1-liter glass beaker is filled to the brim with olive oil at room temperature. The beaker is placed on a range and the temperature of the oil and beaker increases by 25 C°. As a result, 0.0167 liters of...

Q = mSi * CSi * (333.15 K - HeatLost) Q = mHg * CHg * (HeatGained - 293.15 K) mSi * CSi * (333.15 K - HeatLost) = mHg * CHg * (HeatGained - 293.15 K) (3.00 Kg) * (711.62 J/(Kg⋅K)) * (333.15 K - x) = (6.00 Kg) * (138.138 J/(Kg⋅K)) * (x - 293.15 K) (2134.86 J/K) * (333.15 K - x) = (828.828 J/K) *...

I assume it is (20°C + 60°C) / 2 = 40°C (313.15 K), being that no energy left the system in the process.

Homework Statement A 3.00-kg block of silicon at 60.0°C is immersed in 6.00 kg of mercury at 20.0°C. What is the entropy increase of this system as it moves to equilibrium? The specific heat of silicon is 0.17 cal/(g·K) and the specific heat of mercury is 0.033 cal/(g·K). Homework Equations Q...

My bad, I calculated it wrong. I tried again and got the right answer. Thank you!

I did what you said: I squared the velocities, then I calculated the weighted average (same way as I did above), and then took the square root of the weighted average. I ended up getting 82914.03 m/s, which is close to the answer I originally got but not the correct answer. What am I missing?