Recent content by Abdeln

Thank you very much for your help.

Oh so I would use the formula for PE which is mgh, and since its at the top of its swing the PE is the TE so TE = mgh, then isolate for h, convert that to L(1-cos(θ)) and have the new equation L(1-cos(θ))= TE/mg, then further isolate the cos(θ) to cos(θ) = -1 + TE/MGL. then i get the angular...

can I simply do L(1-cos(θ)) ?

Got the first part, and period is 2piSQRT(L/g) so no it doesn't. Is how high the pendulum can swing simply the amplitude of the pendulum?

Homework Statement A pendulum with a mass of 10kg and a length of 4.1 meters is pulled back 26 centimeters to the right from the vertical and released. As it is released, an additional amount of work equal to 5x10^-1 joules is done on the pendulum in the tangential direction toward the left...

I was unsure--I managed to solve it.

Would I now plug everything I have into the projectile motion equations?

ok i changed the magnitude to m/sec to make it easier so it'd be 41.67 m/sec, that'd mean that the x direction it would equal 40.8 m/sec and the y direction would equal 8.3 m/sec. right?

knowing that the magnitude of velocity is 150 km/hour?

Oh, just you trigonometric functions and you get the y direction to be 10, and the x direction to be 49?

Wouldn't it be the SQRT(Vx^2+Vy^2) ? Using Pythagorean theorem?

I believe so, yes

I don't know. I was never taught that--he gave us a problem set and gave us 2 days to solve it.

I found the angle to be 11.53 degrees, what can I do from there?

Homework Statement (Assume no friction) A car that has a mass of 3970 kg is accelerated from an initial velocity of 0 up a ramp that is 50 meters long (the hypotenuse) and has a height above the ground at the end of the ramp of 10 meters. The velocity of the car at the end of the ramp is 150...