Recent content by poopcaboose

Oh a unit step therefore you must translate both H(t) and F(t). LH(t-1)=e^s and f(t)=t^2, L(t^2)=2/s^3, so the transform is 2e^s/s^3

Wouldn't the quarter turn to the stable position in air therefore making the cross sectional area equivalent to the diameter times the width of the quarter?

Of course you are really doing the same thing to both sides. But a short cut is to cancel the operation and do the opposite operation to the other side. This always gives you the right sign and takes less thought and time.

That is the theory of relativity, time changes not velocity.

Does mass also have an effect on time? can't gravity warp the reference of time? If this is true the rest reference of the rocket will be insignificant compared to the rest reference of the earth.

If the original function is e^t*t^2 the translation = F(s-1) and since f(t)=t^2 its transform is 2/s^3, therefore you substitute s-1 in for s and get 2/(s-1)^3. Unless you are doing an inverse translation on the t-axis.

Well you know that the tension will oppose the centripetal acceleration and that the centripetal force will equal the tension. Therefore you are close T=m(w^2*r) where w means omega, or angular velocity, and r is the radius. Just don't square the radius it gets canceled out in the math.

I think amu is the reciprocal of avogadro's number then converted to kilograms, avogadro's number is essentially a conversion factor from atomic mass units to grams. Try converting the amu constant to grams then taking the reciprocal. Then look up avagodros number. So really they can by...

That notation makes no sense, are you trying to translate the function on the s-axis? If you are what is the original function.

I think Iron filings would actually follow the magnetic flux rather than the current.

You have to find the differential equation of the voltage as a function of time. Use laplace transforms of the capacitance value to find impedance ie 1/cs ohms, then use nodal and mesh analysis to find the function. Inverse transform them back into the time domain then you have the voltage...

Sorry about the first post I edited it, you were on the right track just use substitution

Try using the position kinematic equation y=-1/2gt^2+vsin(60)t+yo and set the final value (y)at six meters. And the equation x=-1/2gt^2+vcos(60)t+xo, since there is no acceleration in the x vector the equation becomes 18=vcos(60)t, therefore v=18/cos(60)t sub it into the y position equation...

I have been trying to map the location of the sun in the sky by using three dimensional vector analysis. I can find the rectangular equations of the elliptical orbit with the sun at one of the focal points, however I don't know the function of the angular acceleration of it's orbit. I know the...

Since the 2s orbital of the Be is all filled up and it is bonding to the 2p3 orbital in the F two sp hybrid orbitals are created to allow the bond to happen. These orbitals are created in place of the 2s and 2p orbitals to allow the bond to occur bases on the pauli excursion principle. only one...