Recent content by Xenon

So u = (0.125)V0/(sinθ) How is that looking? Thank you very much, you have no idea how glad iam, that you are helping me.

I got u=05V0*sin30/2sinθ

Could you please show me how to do it correctly, I tried to do it, bit by bit... and it still not right. Thanks

This is what i got: 0 = -2m*u*sinθ+m*(V0/2)sin30 for u. u=-m*sinθ+0.5v0*sin30 2m*V0+m*0 = 2m*u*cosθ+m*(V0/2)*cos30 solve for cotθ cosθ=u+0.5v0*30 when plugging u into cosθ to solve for cotθ cotθ=-m+0.5v0*sin30+0.5v0*cos30 What now? Thank you for all of your help :)

why does it cancels? umm, iam really confused why does it cancels? we are left with u/2 don't we?

U=-m*sin tetha-sin30/2 ?

My problem is that i just can't simplfy it. could you show me what to do with the parameters? V0 and the θ angles. Thank you for helping me.

Homework Statement Mass of 2m slides on a horizontal plane with the velocity of v0 across the x-axis. And hits the mass m which is in rest. after the collision mass m moves with the velocity of v0/2 in the direction which creates an angle of 30 degrees of the original movmen of 2m. a) What is...

How can i know how much place does each one take if i don't have a refernce point?

Pw+Po=4800 Pa+4000 P10cm level= 8800Pa so, 8800Pa* area at the bottom (is it 12^2)? And what about the pressure at the top, the 2 cm? Thank you, iam really struggling with physics, and you are making it much easier.

Pressure of water P=1*4*1000= 4000 Pa

Pressure of oil at depth of 6cm: P=phg P=0.8*6*1000= 4800 Pa Pressure of water 4cm deeper then that h=10 cm?? P=1*10*1000= 10000 Pa 48*(4800+10000)= Iam getting lost with the units. what do you think so far?

Homework Statement Wooden qube with 12cm faces floating inside water and oil in 2 layers. The oil floats over the water. Water density: 1gr/cm cubed Oil density: 0.8 gr/cm cubed What is the mass of the cube. Homework EquationsThe Attempt at a Solution F(buoyancy) = g*(density)*volume , but...