Recent content by SkiingAlta

12.55 n.

8895.15?

So the other force acting would be the weight? Which would be 5.4 N. Subtract that from the upward force? I'm confused...

Okay, so I use f=ma... F=.55*13 F=7.15 N That right? Now how should I start this?

Hey all. Sorry for asking so many questions about force, but these are tricky to me! Okay so I have 2 questions that I have started, but have yet to complete. Help would be greatly appreciated. 1. A model rocket weighs 5.4 N. a) What is its mass? b) What applied force makes it...

Oh. That makes sense! Thanks so much for your help!

Awesome. Thanks so much for the help! One last one possibly? A brick has a mass of 1.2 kg. A force of 5.4 N just begins to move the brick along the floor with a constant velocity. What is the coefficient of static friction? So I would use f=μn 5.4=μ*11.76 μ=0.46 Would that be right...

And for the latter, F=ma F=75(-4) sumF= -300 Yay?

So for #2 if I were to do: Vf=Vi+at 200=0+a6 200=6a a=33.33333 Then: F=ma F=5000(33.333) F= 166,667 NRight?

Hey everyone. Struggling a little bit with a force test I have tomorrow. I've tried a few problems, and don't even know where to begin on another. I have the answers, and I would like to see if they are correct. Thank you so much! 1. An object of mass 70 kg is accelerated by a net force...

Hi, I am a bit confused with all this friction stuff... How would I begin to solve this problem and by using what formulas? Thanks. Suppose a 65-kg boy and a 45-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the accelleration of the girl toward the boy...

Homework Statement Okay, so this is some trig I learned last year but have since forgotten. If you can give me the first step, I can solve the rest on my own. The given statement is true and you have to prove why using Pythagorean Identities. csc2\alpha-1 = cos2\alpha ________ csc2\alpha

Oh duh. Lol. Thanks... I don't process information well at night. Haha. Thanks for the help, I really appreciate it. G'night to ya!

What is the direction of the velocity? Well I guess it doesn't really matter as long as its constant... i/e, you could call acceleration +9.8 as long as Vi was -2.0... Is that what you mean?

Sweet, so technically, since that is "gravity" withheld, the acceleration would be -9.8?