Recent content by downwithsocks
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Young's interference experiment problem
Sorry it showed up for me. Should work now.- downwithsocks
- Post #3
- Forum: Introductory Physics Homework Help
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Young's interference experiment problem
Homework Statement Figure 35-25 shows two sources S1 and S2 that emit radio waves of wavelength λ in all directions. The sources are exactly in phase and are separated by a distance equal to 1.5λ. The vertical broken line is the perpendicular bisector of the distance between the sources. (a) If...- downwithsocks
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- Experiment Interference
- Replies: 2
- Forum: Introductory Physics Homework Help
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Current and electric field strength question
V = IR? Wouldn't I need a value for the resistivity then?- downwithsocks
- Post #5
- Forum: Introductory Physics Homework Help
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Current and electric field strength question
I'm not sure how to tell which section has the greatest voltage drop...looking in my book I can't find an equation that relates voltage and area..- downwithsocks
- Post #3
- Forum: Introductory Physics Homework Help
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Current and electric field strength question
Homework Statement A wire carries a current and consists of three sections with different radii. Rank the sections according to the following quantities, greatest first: (a) current, (b) magnitude of current density, and (c) magnitude of electric field. Homework Equations i =...- downwithsocks
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- Current Electric Electric field Electric field strength Field Field strength Strength
- Replies: 5
- Forum: Introductory Physics Homework Help
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Solving Arc Length of 9x^2 = 4y^2 from 0 to 1
Yeah I get the same answer, thank you!- downwithsocks
- Post #8
- Forum: Calculus and Beyond Homework Help
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Solving Arc Length of 9x^2 = 4y^2 from 0 to 1
Oh so I do the integral from 0 to 2/3? And then get 2(sqrt(13)/3) because the "curve" is in quadrant 1 and 2?- downwithsocks
- Post #6
- Forum: Calculus and Beyond Homework Help
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Solving Arc Length of 9x^2 = 4y^2 from 0 to 1
Times 2 since in the second quadrant 0 <= y <= 1 is still true...so sqrt(13)?- downwithsocks
- Post #4
- Forum: Calculus and Beyond Homework Help
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Solving Arc Length of 9x^2 = 4y^2 from 0 to 1
Oh...okay I guess that makes sense, it means like the set of real numbers in two dimensions? So if I just treat the curve as two line segments.. 9x^2 - 4y^2 = 0 18x - 8y(y') = 0 -8y(y') = -18x y' = 9x/4y (y')^2 = 81x^2 / 16y^2 1 + (y')^2 = 1 + 81x^2 / 16y^2 = (16y^2 + 81x^2)/16y^2 = (16y^2 +...- downwithsocks
- Post #3
- Forum: Calculus and Beyond Homework Help
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Solving Arc Length of 9x^2 = 4y^2 from 0 to 1
Homework Statement Homework Equations l = int( sqrt( 1 + (dy/dx)^2) dx) from a to b The Attempt at a Solution So far I'm stuck at the R^2 thing. I know if it was just R it would mean the set of all real numbers, but I'm not sure as to what R^2 means and I don't know how to google...- downwithsocks
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- Arc Arc length Length
- Replies: 7
- Forum: Calculus and Beyond Homework Help
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Charging Capacitors: Find Change in Charge
Thanks so much, I was confused because I thought the C values were charges in coulombs, so I didn't know how to use my knowledge of voltage given only charges.- downwithsocks
- Post #8
- Forum: Introductory Physics Homework Help
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Charging Capacitors: Find Change in Charge
So then the first one won't change, the second one will increase, and the third one will decrease?- downwithsocks
- Post #6
- Forum: Introductory Physics Homework Help
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Charging Capacitors: Find Change in Charge
Are 6q and 3q capacitances then? I feel really stupid that I just can't work out how to solve this.- downwithsocks
- Post #4
- Forum: Introductory Physics Homework Help
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Charging Capacitors: Find Change in Charge
Homework Statement The figure shows three circuits, each consisting of a switch and two capacitors, initially charged as indicated (top plate positive). After the switches have been closed, in which circuit (if any) will the charge on the left-hand capacitor (a) increase, (b) decrease, and (c)...- downwithsocks
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- Capacitors Charging
- Replies: 7
- Forum: Introductory Physics Homework Help
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Tension in a pendulum (probably just me being stupid)
Well, not equal, since mv^2/r isn't 0, but you get the idea...I'm a little tired.- downwithsocks
- Post #15
- Forum: Introductory Physics Homework Help