Recent content by Mowgli

Trying to figure this out: b.) (r/R)^2 so, therefore it should be (1.75/10.5)^2 which = .0278 c.) MA= F/f = 18.5/.514 = 35.9 ?

(18.5/514)^2 ?? and then I think c.) is MA=F/f = 18.5/.514 = 35.9 Can anyone help?

I just realized my problem had the wrong numbers- so I edited the problem... For part b.) is my ratio now 18.5/514?

14.5/101

is it 1/10?

Right, that makes sense- thank you so much!

I'm not sure how to find that?

I'm sorry- I am not good at this. Are you saying the lift is the same as the weight?

Homework Statement Air (density= 1.29 kg/m^3) passes over the top surface of an aircraft whoe effective wing surface area is 75.5 m^2 at a speed of 255 m/s. If the air speed beneath the wings is 135 /s, what is the weight of the plane. Homework Equations W=mg The Attempt at a...

Homework Statement In a hydraulic lift the radii of the pistons are 1.75cm and 10.5cm. A car weighing 18.5kN is to be lifted by the force of the large piston. a.) what force must be applied to the small piston? b.) when the small piston is pushed 1.00cm, how far is the car lifted? c.) find...

Is this correct? mass m = 0.25 Kg diameter d = 31 cm radius r = 15.5 cm = 0.155 m moment of inertia I = (1/2)mr^2 = 0.003 Kg m^2 (a) ω1 = 0 n = 33 1/3 = 100/3 = 33.33 rpm = 0.5555 rps ω2 = 2πn = 2*3.14*0.5555 = 3.48854 rad/s t = 1.5 s we know that torque τ = I α = I (ω2 - ω1)/t = 0.00697708 N...

what else needs to be done?

Homework Statement An old 33 1/3 rpm turntable may be approximated by a cylindrical disk of mass 0.250kg and a diameter of 31.0cm. a.) What average torque will bring it up to operating speed in 1.50 seconds starting from rest? b.) what net work is done on the disk in bringing it up...

so 1/2(3.5 + 0) (15) = 1/2 (52.5) = 26.25 rev

then 0=175(final velocity) + 40 final velocity = -40/175 = -0.23 m/s ??