Recent content by Shahar

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    Time of a falling object when the force of gravity isn't constant

    Yes, ##M## remains stationary relative to ##m##. I would be happy to derive an equation for ##M## ~ ##m## so ##M## isn't stationary, but I don't have enough knolige in calculus.
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    Engineering Railway engineering requirements

    Just wanted to say that the speed record set by the JR L0 maglev train is an example to why railway engineering is great.
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    Time of a falling object when the force of gravity isn't constant

    I learned a lot from this question and I think, with a bit more calculus I will be able to find the motion equations with air resistance
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    Time of a falling object when the force of gravity isn't constant

    This way is much clearer. I think Ill learn more calculus and then try to get to get to this equation. Or try another simular one( I want to calculate V and t for a free falling object with air resistance )
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    Time of a falling object when the force of gravity isn't constant

    I recalculated θ. θ = 0.78 So t = 2528 seconds.
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    Time of a falling object when the force of gravity isn't constant

    r0 is 2*10^7. I made a mistake when I misread the question. And θ1 is only 0.00031 not 10-7
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    Time of a falling object when the force of gravity isn't constant

    Ok, I read post #50 a few times and I am lost. I don't understand how the equations contributor to the solution.
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    Time of a falling object when the force of gravity isn't constant

    a = G * M v(x) = √(2 *a)/x 1/v(x) = 1/√(2 *a)/x = √x / √2a ∫ 1/v(x) dx = =∫ √x / √2a dx = =1/√2a ∫ √x = =1/√2a * x1.5/1.5 = = √2 * x1.5/3a I'm not sure it's correct. And I reread post #27 about this equation but I don't undersatnd how to apply it. From what I understand this is t, and It...
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    Time of a falling object when the force of gravity isn't constant

    I integrated 1/v(r) dr (and v(r) )and I had an expression that I cna't understand what it means.
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    Time of a falling object when the force of gravity isn't constant

    I got 2000 seconds, which is the only answer that is in the limit 1414 < t < 2828 seconds. I found that the average accelerationis 5 m/s. a0 = 2.5 m/s2 and af = 10 m/s2. I rejected this idea twice already but becuse it's the on;y one that gives an answer in the limit I tried it again. The...
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    Engineering Railway engineering requirements

    I'm a 10th grade student and I'm very interested in railway engineering. Especially railway vehicle engineering. What are the requirements for that kind of profession?
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    Time of a falling object when the force of gravity isn't constant

    Using conversation of energy I found V(r) => Vfinal. And I know the function a(r) = G × M / r2. So when V0 = 0 ,∫ a(r) dr = (Vfinal * Δt)*(Δt/Vavg) = Vfinal * Vavg. Δt = Δx/Vavg.
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    Time of a falling object when the force of gravity isn't constant

    I used averages(V , a) . If I just combine all of the steps into one big expression, would it be an X(t) equation?
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    Time of a falling object when the force of gravity isn't constant

    I found a way to get the time.But because I have to learn more calculus I can't derive an X(t) function. I'm learning calculus alone amd when I will understand how to solve a deferential equation Ill try to derive an X(t) function. So I think rcgldr can release his analysis now.
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    Time of a falling object when the force of gravity isn't constant

    Yeah, its impossible to get v(t). I'm now just searching for x(t). I know there are a lot of solutions here but I don't really understand them. I think I have to get to them myself to fully understand this.
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