Recent content by 1729

  1. 1729

    How Does the Charge Q Affect Forces Between Q1 and Q2?

    I understand now. Since magnitude is irrespective of direction, we shouldn't consider the vector's direction when talking about its magnitude. Hence: $$|k\frac{|Q_1||Q_2|}{2r^2}|=|k\frac{|Q_2|}{r^2}|\cdot||Q|-\frac{|Q_1|}{4}|\\ \Leftrightarrow 0 \ \mathrm{N} =...
  2. 1729

    How Does the Charge Q Affect Forces Between Q1 and Q2?

    It's eastbound? I think I don't understand your analogy. The magnitude would very slightly decrease. Q_1 and Q_2 attract each other heavily, whilst Q and Q_2 would repel each other ever so slightly.
  3. 1729

    How Does the Charge Q Affect Forces Between Q1 and Q2?

    I believe so. We are multiplying a vector by a number, by doubling the magnitude the force vector would double along with it.
  4. 1729

    How Does the Charge Q Affect Forces Between Q1 and Q2?

    We are working with vectors that lie on the same line, but point into opposite directions. I've more or less defined the unit vector to point from Q_1 to Q_2, from left to right, that is.
  5. 1729

    How Does the Charge Q Affect Forces Between Q1 and Q2?

    Homework Statement A positive point charge Q_1 (4.0 µC) and a negative point charge Q_2 are placed near each other. When a negative charge Q is placed in the middle between the two point charges, the magnitude of the resultant forces on Q_1 and Q_2 doubles. What is the charge of Q? (answer...
  6. 1729

    Problem involving gas laws and hydrostatics

    You are absolutely right. There is no way it can rise above the water surface. Now the problem actually makes sense to me; I can see what the gauge pressure formula is doing. The pressure at the water surface inside the tube is equal to 1000 Pa, therefore you can use the formula on the basin...
  7. 1729

    Problem involving gas laws and hydrostatics

    A column of approximately one metre. (using gauge pressure formula) Where does this metre start however? Does it start at the very end of the submerged part of the tube, up the tube to the very top of the water column? I'm absolutely certain it doesn't sink. I double-checked when I translated...
  8. 1729

    Problem involving gas laws and hydrostatics

    Homework Statement A glass tube filled with air at room temperature is 1.54m long. The tube is closed on one end, and open on the other. When submerging the open end in water, the water in the tube rises by 0.14m. How much of the tube is above the water's surface? The correct answer is 0.40m...
  9. 1729

    Simplifying SI Units: What Quantity Does JNW⁻¹kg⁻¹ Denote?

    Velocity it is. Thanks for the help everyone! I was never taught what the exact definition of the Newton was.
  10. 1729

    Simplifying SI Units: What Quantity Does JNW⁻¹kg⁻¹ Denote?

    Homework Statement What physical quantity does ##\mathrm{J\cdot N\cdot W^{-1}\cdot kg^{-1}}## denote? Homework Equations You can simplify using mechanics equations or ones from electrodynamics. Be creative, I guess. The Attempt at a Solution ##\mathrm{J\cdot N\cdot W^{-1}\cdot kg^{-1}}##...
  11. 1729

    Problem involving Coulomb's law

    That's beyond the scope of the course I'm taking, but interesting nevertheless. Could you expand on why that happens?
  12. 1729

    Problem involving Coulomb's law

    Thank you for your answers! I see the error of my ways now -- your argument is convincing, DrClaude! At first I was skeptical about the relation between the electric field property Sammy mentioned and Coulomb's law. Luckily, it then struck me that one can also define Coulomb's law in terms of...
  13. 1729

    Problem involving Coulomb's law

    Homework Statement A big metal sphere has radius ##r## and charge ##Q##. The distance between a certain point charge and the surface of the sphere is equal to ##s##. The magnitude of the force exerted on the point charge is inversely proportional to: A) ##s^2## B) ##r^2## C) ##(s+r)^2##...
  14. 1729

    Determining the mass of a free to move piston

    That's a very interesting way to think about it, it's new to me. It's a lot faster than what I did too! I have a few questions about this method that I'll ask tomorrow hopefully (it's 2.30am right now). Just wanted to let you know I figured it out and it worked. Thanks!
  15. 1729

    Determining the mass of a free to move piston

    Thank you for your reply -- this is a very satisfying explanation! I hadn't thought of the atmospheric pressure exerting a force on the piston as well. Just for the record: 308.54 N - 1.01*10^5 Pa * 0.0027 m2 = mg --> m = 3.65 kg (I'm wondering because of the slight difference with the answer)...
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