Recent content by nightshade123

Homework Statement u(t) * u(t) * u(t) * indicates convolution Homework Equations i know u(t) * u(t) = t u(t) The Attempt at a Solution so (t u(t)) * u(t) = \int \tau d \tau limits of integration are 0 to t so the answer is 1/2 t^2 u(t)?

wouldnt v or h be given in the question? these are just general questions i have right now not h/w or anything. but for velocity (mv^2)/r = m*g v^2 = g*rim not to sure what to do for height

anyone?

1)A mass 'm' is dropped from a spring with constant 'k'. find the time it takes to reach equilibrium. im pretty sure i can use this eqn T = 2*PI*sqrt(m/k) 2)find the tension at the lowest point of the pendulum, with length L and mass M. there will be zero work done by...

you don't have to do all that work A = x work it out yourself. A = .065 m thanks for the help tho

1/2kx + 1/2mv = 1/2kA kx + mv = kA v = 0 kx = kA x=A F = -k * x x = - F/k F = mg x = (mg)/k x = A

at the highest and lowest points v = 0 and Potential Energy is at its max and kinetic energy is 0 i know i have this eqns E = U + K U = 1/2 K A^2 (cos(omega t ))^2 K = 1/2 K A^2 (sin(omega t ))^2 E = 1/2 K A^2

for part b) omega = sqrt (k / m) freq = omega / (2PI) T = 1/freq i can't figure out how to find amplitude

Homework Statement A massless spring of spring constant k = 74 N/m is hanging from the ceiling. a 490 g mass is hooked onto the unstretched spring and allowed to drop. A) Find the amplitude B) Find the period of the resulting motion Homework Equations f = 1/T F=-kx x(t) = Acos...

that makes sense! and it produces the 17.962 rad/s instantly! thanks for the help doc al, since i forgot to change accel to negative i showed how to produce the same answer, but with a lot more work, lol and i was just throwing out that idea for the 2nd part, i knew it didnt make sense but it...

im having trouble interpreting my answer \omega0 = 24.1 rad/s asuming that is right \omega = \omega0 + \alpha * t then i put it in this eqn to find \omega = 30.238 rad/s so next i would take \omega - \omega0 to find the change over 3.1 sec? thus producing an answer of 6.138 rad/s...

[SOLVED] Angular Speed \tau = r * F Homework Statement youve got our bicycle upside-down for repairs with its 66 cm diameter wheel spinning freely at 230 rpm. the mass of the wheel is 1.9 kg and is concentrated mostly at the rim. you hold a wrench against the tire for 3.1s with a normal...

Doc Al, Since they have equal mass shape and such... can't you just say that the cue ball will transfer all the momentum to both balls equally resulting in the two balls to have final speeds of 1/2 v_0? which means the cue ball final speed would be at rest.. Case 2: m_1 = m_2 thanks.

can u say ( also have this problem, and this is how i was working it out) m_ast*v_ast*cos(200) + m_tank*v_tank + m_cam*v_cam*cos(20) = 0 these two equation actualy equal each other (produce same answer) -mavax=mcvcamx+mtankvtankx all withrespect to the x axis can someone...

i think my answers now are correct, and the negitive on the 1/3 is for direction, doesn't mean that momentum is not being conserved