Recent content by Sleepy_time

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    Mixed parallel & series RLC circuit

    You're welcome. I'm glad I could help you with the problem in a different method to what you normally do. If you're finding any difficulty with my method just ask any questions you want to.
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    Mixed parallel & series RLC circuit

    Hi mrmlica. Not too sure what you're trying to do, but the equivalent impedance of a parallel circuit is just: \frac{1}{Z_p}=\frac{1}{Z_1}+\frac{1}{Z_2}+... and for series is: Z_s=Z_1+Z_2+.... Where Z is the complex impedance, Z_n is the n'th component in the given circuit, s=series &...
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    Electromagnetism question involving two stationary conductors and crossbars

    Hi mrcheeses. The question says that "the bar is released and fall", so is the bar under the influence of gravity or not? Also for this situation you should use F_B=BIL for the force on the bar due to the magnetic field. You can find the current flowing through the bar using Faraday's law.
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    Year 12: Cambridge Physics Problem (Oscillatory Motion)

    I suggested this method for a way of figuring the forces on any mass due to a spring regardless of which mode/modes it is in. Because from this it gives a general 2nd order differential equation of motion for the mass, in solving, integration constants will appear which are determined by the...
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    Stone thrown from a building

    No not required, just a preference
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    Stone thrown from a building

    Hi professordad. If you're confident in using vectors and integrating, for this problem you can use \mathbf{F}=m\mathbf{\ddot{r}}=m(0,-g,0), separate the components and integrate along with the boundary conditions \mathbf{x}_0=(0,h,0), \mathbf{v}_0=(v_0cos(\theta),v_0sin(\theta),0). After...
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    Prove square of four-momentum is relativistic invariant

    Hi Falken_47. If you want to derive it explicitly, take the square of the 4-momentum p^i=({\gamma}mc,{\gamma}m{\mathbf{v}}), apply the Lorentz transformations for a contravariant 4-vector to obtain p^i{'} which moves in a frame at an arbitrary speed V with respect to the original frame. So you...
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    Year 12: Cambridge Physics Problem (Oscillatory Motion)

    Hi johnconnor. So if we consider the motion of mass b: the force acting on b will be m\ddot{x}_b=k{\Delta}x_{ab}+k{\Delta}x_{bc}. You can find the extension and direction of the force by considering displacing a slightly, and displacing b a very large distance in the same direction, the...
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    Calculating Power Factor in and RLC Circuit

    Hi rott3np3anut. The Power factor for an AC circuit is just cos\phi. Also tan\phi=\frac{X_L+X_C}{R}.
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    Universal Gravitation and Gravitational Field Question

    Yes I believe it is. Just careful with rounding your answer. Rounding 4.48446... will give 4.5N (1d.p.) or 4.48N (2d.p.)
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    Moment of inertia of a three particle system

    I got the answer that you had as well. Make sure that you've read the question correctly.
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    Universal Gravitation and Gravitational Field Question

    Hi EE123. The gravitational force due to the Earth seems to be correct. But for the moon, the equation F=\frac{Gm_1m_2}{r^2} where r is the separation of mass 1 and mass 2. The rocket is between the Earth and the Moon and you know the distance from the Earth to the rocket and from the Earth to...
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    Is static frinction a conservative force?

    Hi Stergios. Kinetic friction is non conservative because it dissipates energy as heat from the system to the surrounding. So does static friction dissipate any energy from the system and to the surroundings?
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    Magnetic field of two current-carrying wires crossing over

    Hi ness87. The distance from both wires at P are equal and the fields due to both are the sum of their individual fields. So the field would be B=2\frac{\mu_0I}{2{\pi}r}. Using B=6.8\times10^{-3} T also, gives you a factor of 2000 off from the answer
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    Solving for Car & Biker's Meeting Point w/ Variable d

    Ok so: x_{bike}=-d+v\frac{v}{a}=-d+\frac{v^2}{a} You then have d=\frac{v^2}{2a} \rightarrow x=-d+2d=d, you chose your coordinates such that the intersection is at x=0, so they'll meet a distance d after the intersection. I hope you realize how I've come to this.
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