Recent content by Touran Khan

Alright that's what I was thinking as well, no idea what else it could be. Thanks for the help.

Oh right, my bad. So for vertical forces my equation ends up being Psin10 - 70cos45 + Fn = 0. From the horizontal forces equation I calculated P to be 57.369... Then, Fn = 39.535... To sum it up, P = 57.4 and Fn = 39.54. These values satisfy both equations, though do you have any idea what...

Yeah, I'm almost certain that's the issue. I think it's the normal force, but what is the value of that force in this case?

Homework Statement Homework Equations F = ma The Attempt at a Solution My value for P does not make any sense. Why is this so?

That is the diagram I made. Tension and Fg are in opposite directions which is how I got Tcos18 - 68.6 = ma

Homework Statement A 7kg mass is hung from a 1.5 m long massless string. It is released from rest with the string displaced by an angle of 18 degrees from vertical. a) Make a free body diagram of the forces on the mass just after it is released. b) Find the tension in the string. c) Find the...

Ah, apologies for the confusion. Also, which sign error have I made?

How come we need the horizontal ux to find the max height of the projectile?

Oh, alright. Honestly the wording of this question threw me off as well, but the comma was originally there. The question is exactly how my teacher has written it.

Yeah. My teacher gave us a worksheet full of questions that I presume he's made up, so I assumed that to be horizontal position as well.

Homework Statement A student in the back row of class fires a pea shooter in an attempt to hit the blackboard. The student fires it a distance of 23m from his seat to hit the black board. The pea hits the board after 1.4 seconds at a location 9.3 meters below the ceiling. The pea is fired from...

Actually, just took another look at this. Would you mind explaining how you'd use your method to do this? Particularly d = vavet. I feel like that's the same as 2770 meters/250 km/h, no? As that is what I did initially (albeit using scalar quantities, t = d/s) but that was under the incorrect...

Ah, I see. In that case, Runway A) x = 2770 m u = 0 m/s v = 290 km/h = 80.5 m/s v^2 = u^2 + 2as 80.5^2 = 0 + 2a(2770) a = 1.17 m/s^2 Calculating t: v = u + at 80.5 = 0 + (1.17)(t) t = 68.8 s this is my revised solution. Would this method suffice?

Homework Statement A 747 Jumbo Jet must reach a speed of 290km/h by the end of the runway to lift off. The 15L-33R runway at Toronto International Airport is 2,770 meters long. The main Toronto Island Airport runway 08-26 is 1216 meters long. Assuming the jet has a constant acceleration, for...