Recent content by spin360

haha nevermind i got it... 17.76*sin(22) = 6.65m for the height thanks guys!

I got 17.76 m for the "distance" not sure how to convert to height?

Okay... so then would just be y = 14*2.73 + (1/2)(-5.49)(2.73^2)? Actually yeah that's wrong too

Ahh good catch. Okay then I get -5.49 m/s^2 for the accel and 2.73 for t. I plugged that into the y formula and get -5.12 which is obviously not right :frown:

So the problem states: A 1.4 kg wood block is launched up a wooden ramp that is inclined at a 22 deg angle. The block's initial speed is 14 m/s. Use Uk = 0.20 for the coefficient of kinetic friction for wood on wood. U = mu... (a) What vertical height does the block reach above its...

oh wow what a stupid mistake, thanks! it works now

Not sure where you're getting the cos... both are sine since the x vector is opposite of the angle.. making it sine. Right?

Homework Statement A 1100 kg steel beam is supported by two ropes. Each rope has a maximum sustained tension of 6200 N. Then it shows a diagram of two ropes holding a steal beam at the center, both are angled out. The rope on the left is angled away at 20 deg from y axis. The rope on the...

Homework Statement The position of a particle is given by the function x = ( 8t^3 - 3t^2 + 5) m, where t is in s. (a) At what time or times does vx = 0 m/s? (b) What are the particle's position and acceleration at t1? What are the particle's position and acceleration at t2...

Nevermind, I forgot to add 1m :-p

Ahh okay that makes sense. So I got the acceleration right, just by the derivative. I took the integral of 5t^2... which came out to (5t^3)/3... which gives me 1.667m for position. Webassign says that's incorrect?

Homework Statement A particle moving along the x-axis has its velocity described by the function vx = 5t^2 m/s, where t is in s. Its initial position is x0 = 1 m at t0 = 0 s. At t = 1 s, find the position, velocity, and acceleration of the particle. So from that equation... vx = 5t^2, x0 =...