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    Engineering Im having trouble analyzing this circuit

    Also the leg is grounded to obtain electrical voltage? A small voltage is set up across the electrodes by the hearbeat/pulse . This voltage is sensed by the sensor/electrode and is fed to the input of the amplifier. [From a little more research] How does high pass and low pass filters play a...
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    Engineering Im having trouble analyzing this circuit

    Is this because if we did not do this, the circuit would produce alot of leakage voltage?
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    Engineering Im having trouble analyzing this circuit

    I understand the Driven Right Leg circuit thanks to your explaining to an extent, but why is the requirement of resistors in the places that they are placed necessary? And what values of these resistors would be suitable? High values for some? Low for others? Im not fully understanding how they...
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    Engineering Im having trouble analyzing this circuit

    I'm looking at this ECG circuit in particular and im having trouble analyzing it... The ECG circuit and the driven right leg circuit to cancel out the noise. I am a little bit confused on what exactly the r1, the op amp and r2 do... And the values that are generally given to build...
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    Advanced Engineering Mathematics: Euler Method

    I don't think we're learning about Taylor series, but I just don't understand how we would solve the DE... I can probably apply to Euler's method after solving it...
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    Advanced Engineering Mathematics: Euler Method

    Do 10 steps. Solve the problem exactly. Compute the error (Show all details). The problems says do 10 steps, but 3-4 steps will suffice! Problem: y(prime) = (y-x)^2 y(0) = 0 h = 0.1 I don't understand how to get the exact solution and what to do from there! I know that, f(x,y) =...
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    Calculators Integral program for TI-84?

    Thank you for the reply! What does the fnInt function do? And where can I find it?
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    Calculators Integral program for TI-84?

    Any downloads would work too?
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    Calculators Integral program for TI-84?

    I don't know if this is the right place to post a question like this...but help? Does anyone know of a good program that can be downloaded to a ti84 that can solve integrals? It doesn't have to show the steps...just an answer [that you could check while doing homework...if you don't have...
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    Work and Fluid Force: Calculus II

    Ok! Thank you very much. Happy Fathers Day to you! Hope you enjoy the rest of your day! Thanks so much for your help though. You are great.
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    Work and Fluid Force: Calculus II

    Yes. Sorry I just had a typo A = -5 and B = -1 When you evaluate -5 [lower limit] you get -1225/24 And when you evaluate the -1 [upper limit] you get 7/24 So 7/24 - (-1225/24) You get 154/3 times the weight density. Which is 154/3 * 62.4 = 3203.2 ?
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    Work and Fluid Force: Calculus II

    I still get 154/3(62.4) And then when I multiply that out I get 3203.2 ft(lb)
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    Work and Fluid Force: Calculus II

    Theres no pi in this problem, correct? So the problem is just the limits a = -1 and b = -5 and integrating (7/24)(y^2) + (2y+3) With the water density of 62.4 lb/ft^3 outside of the integral as a constant? Correct?
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    Work and Fluid Force: Calculus II

    Oh! Oh! Oh! Sorry! Sorry! I was looking at the other thread and got mixed up! The limits are -5 to -1! -154/3? I did something wrong?
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    Work and Fluid Force: Calculus II

    Yes. Thats where I got the units from.
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    Work and Fluid Force: Calculus II

    They gave us the units for the water density as lb/ft^3 Ohh. But the units for work would be (ft)(lb) Correct?
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    Work and Fluid Force: Calculus II

    I see! Ok. But with my coordinate system. The integral is (-y)(-7/4y - 7/4) dy from the limits [0,4] So integrating 7/4(y^2) + (7/4)y And get 7/24(y^2) (2y + 3) 154/3 as the final answer?
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    Work and Fluid Force: Calculus II

    Hah. I don't know.. Maybe the decimal? Or something. I was hoping for like a nice perfect rounded answer [because we did a couple in class and they were pretty nice numbers...so I just figured that this one would be nice and perfect too] Haha. 4879.1 lb/ft^3 Thank you! Once...
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    Work and Fluid Force: Calculus II

    I see that it could be -y by looking at the picture and where the coordinate system is set... But I posted this question earlier [again] on the forums...because I actually have to turn this in before 5pm today and I didn't know if you would be online before then or not. And someone else...
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    Fluid Force

    So wait, What is it that im integrating? (5-y)(7-7/4y) dy From the limits [0,4] ?
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    Revolving a tank around the y axis: Work needed to pump water

    Wow. Ok. Thanks! The number just seems really weird! Haha. Thats all! Thanks though! :]
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    Fluid Force

    2. A plate shaped as in the figure [picture.jpg is attached] is submerged vertically in a fluid as indicted. Find the fluid force on the plate if the fluid has weight density 62.4 lb/ft^3 The integral I set up was the limits being -5 to -1 and the integral being (5-y)(-7/4y - 7/4) dy...
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    Revolving a tank around the y axis: Work needed to pump water

    1. A tank is formed by revolving y = 3x^2, x = [0,2] around the y axis is filled to the 4 feet point with water (w = 62.4 lb/ft^3). Find the work necessary to pump the water out of the tank over the top. I got the integral being from a = 0 and b = 4 and integrating [(12y-y^2)/3]dy with w and...
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    Work and Fluid Force: Calculus II

    Wait, After starring at this for a while... Would -y work? -(-1) = 1 feet -(-2) = 2 feet -(-3) = 3 feet -(-4) = 4 feet -(-5) = 5 feet ?!?
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    Work and Fluid Force: Calculus II

    I understand that it is the depth in the y direction and all...but I gotta admit that I still do not know how to figure out the formula. I know this is probably really bad...but I just don't know! I also replied to the other question after I integrated it...could you please check that for...
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    Work and Fluid Force: Calculus II

    So At -1: 1 feet At -2: 2 feet At -3 : 3 feet At -4: 4 feet At -5: 5 feet ?
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    Work and Fluid Force: Calculus II

    I integrated this out and got 2(y^2) - (y^3)/9 And plugged in the limits from 0 to 4 with the pi and 62.4 out in front and came up with an answer of 224/9 pi (62.4) And when I multiplied the 62.4 out it came out to be 1553.066 pi Would this be correct? Just seems bit of a weird...
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    Work and Fluid Force: Calculus II

    The depth from the top of the fluid to the end of the triangle is 5 feet At -1: 4 feet At -2: 3 feet At -3: 2 feet At -4: 1 feet At -5: 0 feet -5-y = formula? -5 - 4 = -1 -5 - 3 = -2 -5 - 2 = -3 -5 - 1 = -4 -5 - 0 = -5 Very nice picture, by the way! Thank you!
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    Work and Fluid Force: Calculus II

    Thank you! You are definitely my hero!
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    Work and Fluid Force: Calculus II

    Would the depth be to the top of the liquid? If thats the case, then when y = -2 then the depth would be 7...and thats not in the picture. y = -3 then 8 But thats obviously not right....right?
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