Recent content by sirfederation

ok the question I have is why is the voltage between the 220 olm resistor and the 330 olm resistor 0 V. All resistors in the image are in series so Is=I1=I2=I3 Is = 10/1020 (v/olm) = .0098 or 9.8 mA V(subscript220)= 220*9.8 mA = 2.157 V (this is the voltage between the DC source and...

Here is my final answer for A. A. I = M(R^2)+N(( m(R^2) )/3) Here is my final answer for B. Irim[/size]= I + (M + nm)(R^2) Irim[/size]= M(R^2)+N(( m(R^2) )/3)+ (M + nm)(R^2)

Ok, I redid this problem a couple of times and came up with a different answer. I forgot to set wi to zero in the equation Of[/size] = 0i[/size] + Wi[/size](T)+.5(A)(T^2) in stage one. So Of[/size]=125.66 rad which changes the final answer to 50.2 revolutions

Moments of inertia Help A wheel is formed from a hoop and n equally spaced spokes extending from the center of the hoop to its rim. The mass of the hoop is M, and the radius of the hoop (and hence the length of each spoke) is R. The mass of each spoke is lower case m. A. Determine the...

The Tub of a washer goes into a spin cycle, starting from rest and gaining angular speed steadily for 8 s, when it is turning at 5 rev/s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in 12s. Through how many...

I am posting the correct answer which is indeed -96 meters or 96 meters (determined by where you set your origin).

Here is my diagram: Please tell me if I left out any contact force and friction. Thx I made a mistake with the friction by putting it on the left; it should be upwards.

A 5 kg block and block X are attached to a rope which passes over a pulley. A 30 N force is applied horizontally to the 5 kg block, keeping it in contact with a rough vertical space. The coefficients of static is .3 and the coefficinents kinetic is .3. The pulley is light and frictionless...

Actually is the vector diagrams just the tangent line to the point 1, 2,3,4 on the racetrack?

So wouldn't a a velocity vectory diagram have delta V, Vi[/size], and Vf[/size].

Here are the Diagrams: If I made a mistake please inform me on what I did wrong

Here we are looking down on a racetrack with straight sections and semicircular ends. A car is going around the track and maintaining constant speed. For each one of the numbered positions: Draw diagrams showing the velocity vector for the car. I will post my diagrams on a separate reply...

so if xxf[/size]-xxi[/size]=(Vi[/size])T 43m-0 = 10(m/s)T 4.3s = T Then I plug it into the equation: yyf[/size]-yyi[/size]= Vyi[/size](T)+(1/2)(Ay[/size])(T)^2 yyf[/size]= 5(4.3)+(1/2)(-9.8)(4.3)^2 yyf[/size] =21.5-90.6 yyf[/size]=-69.1 m I set my origin at the top of the building...

sorry the image is on this page Also here is what I have for the sled. T = F1g[/size](sin25) F1g[/size] = (10 kg)(9.8m/s^2) F1g[/size] = 98 N plug into the equation T = F1g[/size](sin25) T = 98 N(sin25) T = 41.41 N Wouldn't Fxg[/size]= Fg[/size]sin 25 and Fyg[/size] =...

Would the diagram for the elevator and sled look like this: Also the tension (t) should be going downwards, Force (F) should be going upwards ( image should be on page two) I accidently deleted off this page ,and now it will not allow me to post on this page anymore.)