The Official AP Physics B Thread

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101

<p>Pressure is pgh.
Directly proportional to height.
Nothing to do with the area.</p>

102

<p>Conceptually, increases in velocity correspond with decreases in pressure.</p>

103

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<p>What’s the formula establishing that relationship? I believe you, but I’m just curious.</p>

104

<p>Well, I got all of the questions on that sample test, however, my work on number 3 was completely inelegant, does anyone have an elegant solution to that question?</p>

105

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<p>Bernoulli’s Equation establishes the aforementioned relationship.</p>

106

<p>Increase in velocity is DECREASE in pressure… really?
I thought it was increase in pressure using v=sqrt 3RT/M
and pv=nrt</p>

107

<p>1st u find the angle. sin-1(10/20) and it give u 30 degrees. Now 2 find net force, all u do is F = mgsin0
F = (2kg)(10m/s/s)sin(30)</p>

108

<p>I think its P1V1 = P2V2, so we established the fact that V2 is bigger than V1, so I think P1 has to be bigger than P2.</p>

109

<p>^^ Correct me if I am wrong guys.</p>

110

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<p>Well, since it’s an inclined plane, we need the angle.</p>

<p>sinθ = Opposite/Hypotenuse = 10/20 = .5</p>

<p>arcsin(.5) = 30º (arcsin = sin^-1; I didn’t want to write it that way because it looks confusing)</p>

<p>If you draw a free body diagram and break up weight, you’ll note that force normal and weight in the y direction cancel out, so you’re left with the net force being equal to weight in the x direction (Wx). To get Wx, you use sin(30º)*20 = 10 N.</p>

<p>Hope that helps.</p>

111

<p>Hey guys, if I’m averaging 45 correct, 20 wrong, and 5 blank on the MC, is that a decent enough score for a 5? I’m focusing mostly on the FRQ’s, because they’re the easy ones.</p>

112

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<p>That nudges you ahead of the curve, which is 55% correct (you’re at 64%). Good work.</p>

113

<p>Thanks IRJunkie, my method was complete bs =)</p>

114

<p>A magnetic field of 0.1T forces a proton beam of 1.5 mA to move in a circle of radius 0.1 m. The plane of the circle is perpendicular to the magnetic field. which is the best estimate of the speed of a proton in the beam as it moves in the circle?</p>

115

<p>approximately 10^6?</p>

116

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<p>Alright, begin by setting force magnetic equal to force centripetal. So,</p>

<p>qvB = (mv^2)/R</p>

<p>This simplifies to:</p>

<p>qB = (mv)/R</p>

<p>Isolate for v:</p>

<p>(qBR)/m = v</p>

<p>Plug & chug:</p>

<p>v = (1.6e-19<em>.1</em>.1)/1.6e-27 = 1e6 m/s</p>

117

<p>thats correct but how did u get that?</p>

118

<p>guys guys</p>

<p>princeton review(the ap physics B bible!!) says :
“the pressure is lower where the flow speed is greater”
“speed increases as the cross sectional area of the pipe increases” - opposite from what you would expect</p>

<p>google “bernoulli effect” for further explanation</p>

119

<p>thank you!</p>

120

<p>qvB=mv^2/r</p>

<p>after simplifying, we have:</p>

<p>v= 958083 m/s and I approximated that to be 10^6 m/s</p>