Recent content by Blakeasd

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    Issue finding change in voltage across a resistor

    Homework Statement Determine the voltage drop across R1 R1 = 770 , R2 = 630 , and R3 = 470 . The equivalent resistance is 816Ω Homework Equations V = IRThe Attempt at a Solution 1.) I find the total current of the circuit: V = IR -> I = V/R I = 12v/816Ω = 0.015A 2.) I then find the...
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    Coefficient of Friction and Energy

    Ah... so if I were to sum the forces: 0 - (u*30*9.8) = 60 u = .204 It works! If you don't mind me asking another question, can you explain you reasoning for: v2 = vi2 + 2a(x - xi) 16 = 4 + 2*a*3 Why don't you use 2m/s for the final velocity? Thanks!
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    Coefficient of Friction and Energy

    I don't understand how I'm using the same force twice. I take the force pushing the object forward and subtract the force of friction pushing in the other direction. When summing forces I thought it is necessary to set their sum equal to ma. Am I missing something fundamental? Thanks!
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    Coefficient of Friction and Energy

    Ok, I see why it is 180J, thanks! Here is my revamped work: W = fd 180J = f * 3m f = 60N 60N - (u * (30*9.8)) = (2/3) * 30 u = -.14 This is not a choice, though: a.) 0.07 b.) 0.12 c.) 0.20 d.) 0.27 e.) 0.60 When I use W = Fd and F = umg , I get .204 (an answer choice). So why doesn't this...
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    Coefficient of Friction and Energy

    Thank you for your response. When I sum up the forces, I set them equal to ma. I need acceleration so I can solve for the coefficient of friction. Can you please explain how you found work to be 180J . Thanks!
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    Coefficient of Friction and Energy

    Homework Statement A 30kg sled is sliding on a frictionless sheet of ice at a velocity of 4m/s. The sled encounters a rough patch of ice and begins to slow down. After traveling on the rough patch of ice for 3m, the sled's velocity is 2m/s. Determine the coefficient of friction between the...
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    Why is one Tension Negative and the Other Positive - Atwood Machines

    Yes, but isn't mass 1 accelerating downwards as well. It also is being affected by gravity. Why then is it equal to positive ma and not negative ma. I expect it to be negative, but that is not the case. Am I missing something obvious? Thanks!
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    Why is one Tension Negative and the Other Positive - Atwood Machines

    I see, thanks. But why is T - m2g = to a negative m2a ? What is the reasoning behind m2a being negative. If when dealing with the first mass ma is positive; what makes the difference here?
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    Why is one Tension Negative and the Other Positive - Atwood Machines

    Why is one Tension Negative and the Other Positive -- Atwood Machines Hello, Please consider the following: I am trying to understand how to solve for acceleration when dealing with Atwood machines. I keep getting hung up on one part: the direction of tension. In the image above, why is...
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    Perpetual Battery with Pickup Coil and Magnents

    Hello, I recently was thinking about an idea for a perpetual battery. I have an extremely limited amount of Physics knowledge (just a Conceptual Physics [Algebraic Physics] class) as I am still in high school. Does anyone know if this would work: 1. Have a tube 2. Wrap the tube in copper...
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