Recent content by frigid

A 2000 kg car is raised 1.5 meters high on a hydraulic jack, which has a large piston with a diameter of 30cm. The small piston has a diameter of 5cm (btw the reservoir is filled with oil) d) When the car is raised by 1.5 m, what distance will the small piston move? Is it simply that you have...

Ok so Fn/Fg = cos(30deg) = 0.866 0.866 * 30kg * 9.8 = 254.6N 254.6 / 0.008m^2 = 3.2 X 10^4 Like this?

But what happens when the object is on an inclined surface? In this particuliar situation a 30kg block with an area of force of 80cm squared is standing on a surface which 30 degrees inclined relative to the floor. How do you calculate the pressure it exerts?

Ff: 0.2*9.8 = 1.96N Fn: 1.96 / 0.3 = 6.53N Fa - Ff = ma 6.53 - 1.96 = 0.1*a 4.57 = 0.1*a a = 45.7m/s^2 The concern i have is that i didn't use the kart's 10kg no where in my calculations. Is this kosher?

Sorry guys, i failed to mention: the front of the cart is flat and vertical, it's not a slope like I'm guessing you assumed it was. So i find the weight of the block: 0.2*9.8=1.96N Fn = 1.96*0.3 = 0.59N How do I find the acceleration? I don't have a velocity :rolleyes:

Maybe it's just me, but i would classify this one as hardcore... Due to a cart's sudden acceleration, a 200 g wooden block remains on the front of the cart and does not slide or fall off. The mass of the cart is 10 kg. The coefficient of static friction between the block and the front of...

Can anyone please help me with this one real quick. I know that Fnet is Fa - Ff = MA His mass is 70kg and velocity 2m/s. But since i don't have the distance of his fall or the time i can't calculate the acceleration. So how do I figure out the magnitude of the air's frictional force which...

A skydiver jumps from an airplane and descends at a constant velocity of 2m/s. The mass of the skydiver and his equipment is 70kg. Calculate the magnitude of the air's frictional force acting on the skydiver and his equipment. F = 70kg x 9,8N/kg = 686N I'm lost