Recent content by Gr33nMachine
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[Thermodynamics] Impossible Heat Transfer Problem
An insulated Thermos contains 122 g of water at 91°C. You put in a 13.7 g ice cube at 0°C to form a system of ice + original water. Use mc(ΔT) |water + mc(ΔT) |ice + Lm |ice = 0 Final T = [(m*c*T) |water + (m*c*T) |ice - (L*m) |ice] / [m*c |water + m*c |ice] Final T =...- Gr33nMachine
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- Heat Heat transfer Impossible Thermodynamics
- Replies: 2
- Forum: Introductory Physics Homework Help
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[Non-uniform Air Density] Pressure problem
I'm really confused. Why are there two boundary conditions for an single variable integration? I know z and h are the same parameter. If I set P=0 and z=h=L in the final equation, I get L = 1.3L2/-1.3L, which simplifies to -1.3L2=1.3L2 How does this help?- Gr33nMachine
- Post #7
- Forum: Introductory Physics Homework Help
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[Non-uniform Air Density] Pressure problem
Well, dP/dz = ρg can be rearranged to dP = gρ(z)dz and integrate ρ(z) from 0 to h? The result is P = g(-1.3/(2L)*h2+1.3h), and now I can solve for L = (1.3h2)/(2P/g-1.3h) But what I can plug in for h?- Gr33nMachine
- Post #5
- Forum: Introductory Physics Homework Help
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[Non-uniform Air Density] Pressure problem
I don't understand how to express the density as a function of z... y = mx + b ρ(z) = -1.3/(z max) * z + 1.3 ? How does this integrate into anything useful?- Gr33nMachine
- Post #3
- Forum: Introductory Physics Homework Help
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[Non-uniform Air Density] Pressure problem
Homework Statement Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3. What would be the height of the atmosphere if the air density decreased linearly to zero with height? Homework Equations P = ρgh The Attempt at a Solution Really not sure...- Gr33nMachine
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- Air Air density Density Pressure
- Replies: 7
- Forum: Introductory Physics Homework Help
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[Point Charges] Can't figure this out
Wow. All I can say is, stay off drugs, kids. edit: So I ended up with a value of 1.016e-22, and it's STILL not right. I've run out of chances now, but I'm still interested in knowing what I'm doing wrong here.- Gr33nMachine
- Post #8
- Forum: Introductory Physics Homework Help
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[Point Charges] Can't figure this out
Ahh... so ƩF = (5F2+F2)1/2 = (26F2)1/2 This still results in a suspiciously large exponent: ƩF = 4.694e-31- Gr33nMachine
- Post #5
- Forum: Introductory Physics Homework Help
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[Point Charges] Can't figure this out
So, ƩFx = F + 8F + 8F/4 and ƩFy = 4F/4 ƩF = (11F^2+F2)1/2 ƩF = (122F2)1/2 F = k(4e)(e)/(.012) = 4ke2*.012 ƩF = 1.0168e-30 This doesn't seem right since it's 7 degrees off my previous calculations.- Gr33nMachine
- Post #3
- Forum: Introductory Physics Homework Help
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[Point Charges] Can't figure this out
Homework Statement In the figure below, six charged particles surround particle 7 at radial distances of either d = 1.0 cm or 2d, as drawn. The charges are q1 = +8e, q2 = +8e, q3 = +e, q4 = +8e, q5 = +8e, q6 = +4e, q7 = +4e, with e = 1.60 10-19 C. What is the magnitude of the net electrostatic...- Gr33nMachine
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- Charges Figure Point charges
- Replies: 7
- Forum: Introductory Physics Homework Help
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[Electric Charge Force] I swear I'm doing this right.
Ok, so I got [436(k^2)(e^4)/(10^-8)]^.5 which comes out to 4.8055e-23. This is still incorrect...- Gr33nMachine
- Post #5
- Forum: Introductory Physics Homework Help
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[Electric Charge Force] I swear I'm doing this right.
D'oh. Pythagorean Theorom then? edit: so, ([20k(e^2)/.0001]^2 + [6k(e^2)/.0001]^2)^.5 then?- Gr33nMachine
- Post #3
- Forum: Introductory Physics Homework Help
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[Electric Charge Force] I swear I'm doing this right.
Homework Statement In the figure below, six charged particles surround particle 7 at radial distances of either d = 1.0 cm or 2d, as drawn. The charges are q1 = +8e, q2 = +8e, q3 = +e, q4 = +8e, q5 = +8e, q6 = +4e, q7 = +4e, with e = 1.60 10-19 C. What is the magnitude of the net electrostatic...- Gr33nMachine
- Thread
- Charge Force
- Replies: 4
- Forum: Introductory Physics Homework Help