Recent content by fernancb

Okay, so the force exerted depends just on the the mass/volume/height of the fluid above it. Not the area. Did I get that right?

Homework Statement The question pertains to hydrostatic lifts. So, assuming you have a lift and you're applying pressure to one side and the other side is rising: P0 + P1 = P0 + P2 + (rho)gh Now, when we look at the equation, why is it (rho)gh? I'm thinking it's something like mgh, so to...

Homework Statement A 1.06 kg hollow ball with a radius of 0.113 m, filled with air, is released from rest at the bottom of a 1.94 m deep pool of water. How high above the water does the ball shoot upward? Neglect all frictional effects, and neglect the ball's motion when it is only partially...

Homework Statement A spring is suspended from a ceiling and has a block attached to its lower end. The block is held a distance y1 from the ceiling (at this point the spring is at its rest length) and released. The block oscillates up and down with its lowest point being 0.091 m below y1. Find...

Homework Statement A 0.222 kg shoe is dropped onto a vertically oriented spring with a spring constant of 101 N/m. The shoe becomes attached to the spring upon contact, and the spring is compressed 0.110 m before coming momentarily to rest. What work is performed by the weight of the shoe...

Okay, so my normal force would then be Fg = N + vertF N= Fg - vertF = (15*9.8) - (74.9sin17.6) = 124 N then my frictional force would be: Ff = (mu)k * N = (0.266) * (124N) = 33.0 N Work done by friction: W = Fd = 33N * 4.91m = 162 J

So the frictional force is: Ff = (mu)k * N = (0.266)* (74.9sin17.6 + (15*9.8)) = 45.1 N But how would I calculate the work done? The block doesn't move in the y-direction, so I would think no work coule be done since W=Fd

Huh? do i use the mass in there somewhere?

Then I took that and put it into W = Fd = (6.02)(4.91) = 29.6 J Still wrong...?

So it'll be : Ff = (0.266)(74.9 sin17.6) = 6.02 N

Homework Statement A 15.0 kg block is dragged over a rough, horizontal surface by a 74.9 N force acting at 17.6 degrees above the horizontal. The block is displaced 4.91 m, and the coefficient of kinetic friction is 0.266. Find the work done by the 74.9 N force = 351 J Find the work done...

Oh no! so my initial velocity would be = 11.8/3.55 = 3.32 m/s

I tried 11.8 m/s and it's wrong

X-axis: (3.55/2)(4.58)(cos25.4)=7.34m/s Y-axis: (3.55/2)(4.58)(sin25.4)+(3.55/2)(3.26) = 9.27 m/s Resolve? So I wouldn't need to bother with angles after? Initial speed: (sqrt(7.34^2)+(9.27^2)) = 11.8 m/s

object moving North p=(m/2)v2 object moving north east p=(m/2)v3