Recent content by boognish

so understanding what you have told me i think that it resolves to: Li= m r^2 w Li = 2.0 x 1.5^2 x 5.4-s Li = 24.3 Lf = Li Lf = (Ii + If) x wf Lf = ((m R^2 + (1/2 m R^2)) x Wf 24.3 = ((2.0 x 1.5^2) + (.5 x 2.0 x 1.5^2)) x wf wf = 3.65s- Any thoughts?

1. A hoop with mass 2.0 kg and radius 1.5 m is rotating with initial angular velocity of 5.4 s-1. A disc with the same mass and radius is dropped on top of the hoop so that the centers coincide. What conservation law applies here? What assumptions must you make to apply it? What is the final...

1. a student throws a ball down with an initial velocity of 5.0 m/s from a point 8.0 m above the base of a track. Find the speed of the ball as it hits the base of the track. The track is designed to losslessly change the direction of the ball up a frictionless ramp. What is the height above...

can anyone confirm if i am correct or incorrect on this?

if P = roh g h 1.013 x 10e5 pa= 1.29kg/m3 x 9.8m/s2 x h h = 8012.97 m?

1. How high would the atmosphere extend (assume P=0 at the top) if it were uniform density Pair - 1.29 kg/m3 and we assumed the air was incompressible (Patm = 1.013 x 10e5 Pa 2. maybe roh g h? [b]3. I don't have any idea where to start with this. There are a few ideas in there...

is this correct..? I am wondering if all the masses should cancel out in this equation or if I have it set up correctly

1. A 3kg block slides down a rough plane inclined at an angle of 27 degrees. If the acceleration of the block is 2.5 m/s/s find the coefficient of kinetic friction. 2. solving for components Force Net = M x A forces in the y direction: Fn - Fg cos 27 degrees = 0 forces in the x...

yea diagram looks like ---------block--------- | | | block

1. A 2.0 kg mass sits on a frictionless horizontal surface. The mass is attached by a weightless string running over a frictionless pulley attached to a 1.5 kg mass. An additional string exerts a force on the block on the mass on the surface so that no motion results. Find the tension in each...

I adjusted the equation to solve for tension as that is what I actually want to know so: Ft x cos 31 + (Ft sin 31/cos 15) x sin 15 - Fg = O When I enter the known values i have this: Ft x .857 + (Ft x .515 / square root of 6 + square root of 2 / 4) x (square root of 6 - square root of 2 /...

so here is where i am at... y: Ft cos 31 + Fc sin 15 - Fg = 0 x: ft sin 31 - fc cos 15 = 0 using substitution i can solve for Fc cos 31 x (Fc cos 15/sin 31) + Fc sin 15 - Fg = 0 when i try to solve for elements of cos 15 and sin I am getting different square roots divided by 4. i don't know...

to answer your question the forces would have to cancel each other out.

ok here is the first part i am stuck at. If i isolate the tension force then i can see that there is a 31 degree angle east of south... the components of y would be Ft cos (31) and x would be Ft sin (31) Where i get confused is the next step shown by my instructor is this equation. y = Ft cos...

1. a 52 Kg mountain climber is suspended from a cliff by a rope. Given a few bits of information the angle formed by the rope to the climber is 31 degrees The angle of the climbers legs is 15 degrees north of horizontal (a) Find the tension in the rope and the force that the mountain...