Recent content by AnthonyC

Oops, my bad, I'm still used to the social studies version of north/south coming first.

I believe that I have answered these questions correctly: Direction of the boat relative ot the river: b) tanθ = (opposite/adjacent) tanθ = (3/4) = 36.9 degrees The boat will be heading NW at an angle of 36.870˚ Boat's Velocity c) a2 = c2 - b2 = √(25 – 9) = 4 m/s Time taken...

Great, thankyou! I wasn't sure which line it should have been on, so I casually put it between the two, haha.

I just figured it was neater to have the entire diagram roated 90degrees clockwise. What do I need to change for it to be correct?

"The river is 50.0 m wide, and flows east at 3 m/s. A boat that can travel on still water at 5 m/s heads in a direction that allows it to cross the river perpendicular to the shore." Based on this wording, this is the diagram that I came up with, though I really am very unsure as to if it is...

I apologize if I don't understand completley, but it should instead be mgsin25 = 2(9.81)sin25 = 8.292 ?

Ok, thankyou, that makes sense to me. So from the object's initial position at the top of the incline, the net force on the object would simple be: N = mg cos25 = 2(9.81)cos25 = 17.78 N Is that correct?

They do not indicate if the horizontal plane is frictionless, they say: "An object of mass m=2.00kg is released from the top of a frictionless inclined plane that makes an angle of 25degrees wiht the horizontal" It hits position X with a velocity of 14.394 m/s, and the acceleration along...

Ok, now I'm stuck at how to find the velocity at position Y. They do not indicate the distance between X and Y. So I know you cannot tell me how to do it, but can somebody point me in the right direction? The acceleration should will decrease while on the horizontal plane, but I'm not sure...

Thankyou! Also, to solve for the kinetic energy at the bottom, can I correctly assume that since it is frictionless, the system is closed, so the law of conservation of mechanical energy would indicate that the Potential Energy at the beginning would be equal to the Kinetic Energy at the...

Meaning having a mgsin(25) coming from the box down the slope, and a mgcos(25) coming perpendicular from the bottom of the box through the slope? (If that makes sense)

I am supposed to take this: http://www.recklesscaution.com/original.jpg and draw the vector diagram of the forces acting on the object. So if it is frictionless, I believe this is all I would need: http://www.recklesscaution.com/vector.jpg Is that correct?

Haha, of course! Wow, I look like a fool. Thankyou!

I have another question. I am again following an example in the book but end up with an incorrect answer. Question: A 125 N box is pulled east along a horizontal surface with a force of 60 N acting at an angle of 42degrees with the horizontal. If the force of friction on the box is 15 N...

Ohhh, I assumed that 180 was required as it was a straight line. There we go, I get the right answers. I knew it must be something very obvious I overlooked. Thankyou once again for all your help!