Recent content by FredericChopin

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    Cut the String - Energy and Friction

    Homework Statement http://imgur.com/FZM5gqC,RlLeGmP#1 http://imgur.com/FZM5gqC,RlLeGmP#0 Homework Equations f^k_{AB} = \mu_k N_{AB} W_{f,i} = \int_{r_i}^{r_f} \vec{F} \cdot d\vec{r} E^{mech}_f = E^{mech}_i + W^{NC}_{f,i} U_{elastic} = \frac{1}{2} k x^2 The Attempt at a Solution Since the...
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    Solve Kinematics Problem: Find θ for sx = symax

    Ahh... I got it. It takes only half the total travel time to reach the top of the trajectory. Thank you.
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    Solve Kinematics Problem: Find θ for sx = symax

    Homework Statement http://imgur.com/BPPbwpu Homework Equations s = ((v + u)*t)/2 - an equation of motion for constant acceleration, which is applicable in this situation. The Attempt at a Solution Let u be the velocity at which the projectile was launched. It is given that sx = symax. So...
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    Friction - 3 Blocks and a Pulley

    O.K., I realized I grossly overcomplicated things. Since the force of static friction on Block 1 due to Block 2 is the only force in the x-direction acting on Block 1, and Block 1 has acceleration a, then Newton's Second Law states that: ∑ Fx = m1*a = fs12 Thus, the force of static friction...
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    Friction - 3 Blocks and a Pulley

    Homework Statement http://imgur.com/C0XYEKw Homework Equations Fnet = m*a FG = m*g fstatic = μ*N The Attempt at a Solution If both Block 1 and Block 2 have acceleration a in the x-direction and have an acceleration of 0 in the y-direction, then the force of static friction on Block 1 due to...
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    Solve Pulley Problem: Frustrated After Weeks of Trying

    Since the workers are pulling with a force f on the rope, then due to Newton's Third Law, the rope would exert a force f on the workers in the opposite direction, so I am missing another 2 f forces, right? So the acceleration should be: a = (4*f - (m + m + 0.5*m)*g)/(m + m + 0.5*m)
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    Quantum physics experiments for EEI

    Hey, are you from Australia? I remember doing E.E.I.s... God I hated them. Well done doing quantum physics as a topic for your E.E.I. though. Complex and challenging, right? That sounds like a hard topic to do not only because the theory is difficult but it is difficult to do an experiment on (I...
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    Solve Pulley Problem: Frustrated After Weeks of Trying

    Homework Statement http://imgur.com/SE50jeP Homework Equations Fnet = m*a FG = m*g The Attempt at a Solution I'm so confused about this problem and I still can't solve it after weeks. The initial distance of the workers is irrelevant. If I treat the workers and the platform as a single...
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    X > 0, but why doesn't 1/x > 0?

    Rather, something like this: If a ≠ 0 and b ≠ 0, then if a = b, then 1/a = 1/b. Well, I'm probably not using the inequality sign correctly and trying to use equality rules for the "greater than" sign. But basically, if x > 0, then a "=" x, and b "=" 0, and using the algebra "rule" above, if x...
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    X > 0, but why doesn't 1/x > 0?

    Oh, well, there is nothing wrong, then. I thought that rule was wrong because if x > 0, then dividing 1 by both sides of the inequality yields 1/x > 1/0. My concern was that 1/0 is not defined. There is clearly something wrong with the algebra, but what is wrong? Thank you.
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    X > 0, but why doesn't 1/x > 0?

    If x approaches 0 from the right, then 1/x will become arbitrarily (positively) large, which is definitely still greater than 0. Hm... I still don't see what could be wrong. Before I posted, I Googled "inequality properties" and I found a popular school study guide called "Sparknotes", which...
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    X > 0, but why doesn't 1/x > 0?

    If a > b, where a > 0 and b > 0, then 1/a < 1/b. But what if b = 0? For example, if x > 0, meaning if x is a positive number, then it should be that 1/x > 0. Yes, yes, I know I would be dividing by 0, but it doesn't make sense intuitively. If x is a positive number, then obviously 1/x is a...
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    General Formula for the Reciprocal of a Sum of Reciprocals

    I know that the reduced mass, μ, of an object is: \mu = \frac{1}{\frac{1}{m_1} + \frac{1}{m_2}} \mu = \frac{m_1 m_2}{ m_1 + m_2 } But is there a general formula (or a simplified expression) for finding the value of: \frac{1}{ \frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n} } ? Thank you.
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