Recent content by Rct33

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\frac{6}{v}=t=2.5γ where the velocity is a fraction of c, 6 is in light years and t, 2.5 are in years. Implies: \frac{6}{2.5}=\frac{v}{\sqrt{1-\frac{v^2}{c^2}}} ∴v=\frac{12c}{\sqrt{25c^2 + 144}}=2.4 solved with wolfram because tired Can't understand why I get 2.4c as an answer?

Homework Statement A spaceship travels from Earth to a star that is 6 light years away. The spaceship takes 2.5 years to reach the star in its frame. Calculate the velocity of the spaceship. Homework Equations x=\frac{x_0}{γ}, t=γt_0 The Attempt at a Solution I guess I have to...

Thank you so much, was a good read :smile:

Hm I am thinking then that there is a limited number of free parameters? So to answer your question, I have no good reason to believe a closed form solution exists. So how would I comment on the number of free parameters? Does this mean anything?

Homework Statement I am trying to find the recursion relation for the coefficients of the series around x=0 for the ODE: y'''+x^2y'+xy=0 The Attempt at a Solution Therefore letting: y=\sum_{m=0}^\infty y_mx^m \therefore y'=\sum_{m=1}^\infty my_mx^{m-1} \therefore...

This makes sense after thinking about it, thanks for your help! It is kinda simple now that you explained it too :p

Homework Statement I am asked to find the radial and transverse velocity and acceleration for a particle with polar coordinates r=e^t and \theta=t The Attempt at a Solution Therefore let the position of the particle be \underline{r}=\underline{\hat{r}}e^t $$\therefore...

is b not equal to -2aV_{2}-2aV_{1}\cos(\alpha)? Thats why I canceled the negative in d2=C-B2/4A

Do you get that result from simplifying a^2+\frac{(2aV_{2}+2aV_{1}\cos(\alpha))^2}{4(V_{2}^2+2V_{2}V{1}\cos( \alpha)+V_{1}^2)}?

Well it still ain't pretty, but I have d, thanks a lot for your help

So t=\frac{aV_{2}+aV_{1}\cos(\alpha)}{V_{2}^2+2V_{2}V{1}\cos(\alpha)+V_{1}^2}? Would I have to plug that into \sqrt{a^2-2aV_{2}t-2aV_{1}t\cos(\alpha)+V_{1}^2t^2+V_{2}^2t^2+2V_{1}V_{2}t^2\cos(\alpha)} to get the distance? Seems awfully complicated

Homework Statement The problem describes an aircraft taking off from a point on a runway with constant speed V_{1}, climbing at a constant angle \alpha, at the point of takeoff, a car drives towards the aircraft a distance a away with speed V_{2}. I simply have to find the closest distance...

V - (R*I1) - R*(I1 - I2) = 0 and V - R*(I2 - I1) - (R*I2) = 0 Therefore I - 2*I1 + I2 = 0 I - 2*I2 + I1 = 0 I solved these to get I1 = I therefore I2 = I Therefore D and C have the same brightness as the original bulb. Will now try for the third circuit. EDIT: Actually...

Okay I will try