Recent content by sweetpete28

Thank you Simon -- I'll give this a go... ".. I don't see how that is "introductory" ... what level is this to be done at?" = My thoughts exactly... ...nd day of Principles of Physics II...

My profound apologies = Yes, in cm. Can you please help a little more...an equation?...Yes, I am really lost on b

1.71 kg is correct. Still lost on b; F = 16.794 N and it is only in y direction

a)Point charge Q = +7.25 μC is fixed at the origin. Point charge q = +1.60 μC is now carefully placed on the positive y-axis, and it floats at (0,7.88). Find the mass of q. b) Refer to question a. Suppose that, as charge q is floating, the point charge Q at the origin begins to lose its...

Pressure, Volume and Temp Change...WORK done? n = 9.95 moles of an ideal gas are slowly heated from initial pressure P1 = .922 atm and initial volume V1 = 679 L to final pressure Pf = 1.16 atm and final volume Vf = 1073 L. Find T1, the initial temperature of the gas. If the plot of this...

Area of the rectangle + Area of the triangle.

Yes -- kilograms, so that is not the issue. Specific question is: Find the amount of water that boils. But below question it also states "Note: reaching the boiling point is not enough, the question asks for the amount of water that vaporizes as well" and to "assume that no water evaporates...

wait wait wait... If we start with .658 kg of water and .148 vaporizes...what is the amount of water that boils?

Are we sure there is a phase change for the lead? If so, is it definitely the full amount?

FYI: Below answer is incorrect. Please help! What am I doing wrong now? 257000 + 303362.8 = 223931.7444 + Q Q = 336.4310556 kJ m = 336.4310556 / 2260 kJ/kg = .148 kg

Ok...what about this: 257000 + 303362.8 = 223931.7444 + Q Q = 336.4310556 kJ m = 336.4310556 / 2260 kJ/kg = .148 kg

.658 Kg is the mass of the water; it's given in the problem. How about this: Energy released when lead turns from liquid to solid = (10.28)(25kJ/Kg) = 257,000J 257000 + (10.28)(130)(327 - T) = (.658)(4186)(T - 18.7) T = 182.2411368 degrees Celsius Heat released by lead = 257000J +...

Evidently not. One last chance to get this right...little help please on needed equation.

Is the specific heat of lead needed to calculate the heat loss of the lead in addition to the energy released from the phase change (change from liquid to solid)?

Do I need to use the specific heat of lead?