# Search results

1. ### Force of Friction Equation?

I think the reason that my problems have arisen is because of my equation for the force of friction. I have Ff= ucosAngle That right?
2. ### Force of Object on Inclined Place

I need extra help. I'm studying for my Final Exam which is tomorrow and I cannot remember how to find forces of objects on an inclined plane. The question I'm having troubles with is: A 12.5 kg crate is placed on a ramp that is at an angle of 22 degrees from the horizontal. Find the...
3. ### Final Exam Question Help

Final Exam Question Help!!! HELP!! I need extra help. I'm studying for my Final Exam which is tomorrow and I cannot remember how to find forces of objects on an inclined plane. The question I'm having troubles with is: A 12.5 kg crate is placed on a ramp that is at an angle of 22 degrees...
4. ### Mechanics of a spring

There is a question involving a simple land rover that can be controlled to go a certain distance. It is two pop bottles connected spout to spout and attached around the middle is a metal tape measure which one would roll up and then let go for the rover to move. I have to explain the mechanics...
5. ### Heat resistant and truly frictionless substance

QUESTION If it were possible to make a tough, heat resistant and truly frictionless substance, what applications would it have? [?]

YES!!!! And I really have to thank you for being patient with my annoyance. I was a little bit annoying and I know it. So thanx. -Paige

Thank you soooooooooooooo much for everything!!! OMG THIS WAS SOOO FRUSTRATING. But thank you sooooo much!! -Paige

How did you get that for acceleration?

v-squared- = vo-squared- + 2aDelta X 4.163 -squared- = 0 + 2(5)a 17.330569 =10a a=1.733

K...I found acceleration wrong. I realized that. Acceleration = 1.733 m/s -squared- So then do I go F=ma F= 10.8 (1.733) = 18.7164 So is that Fnet? If it is...then what?

Ok...so I have that. But what is SOOOOO EXTREMELY FRUSTRATING...IS WHAT DO I DO AFTER I HAVE ALL THAT!! The acceleration becomes 0.4163 m/s-squared-. F= ma.... that makes is 10.8(0.4163)=4.49604 N that's no where near the answer.
12. ### I really need Help SERIOUSLY

Ok....the puzzler. I haven't gotten really far. But this is what I have.... Vertical: Delta X= -12.96 m Vo= 0m/s a= -9.80 m/s-squared- v-squared- = vo-squared- + 2aDelta x v= 15.94 m/s v = vo + at t= 1.626 s Horizontal: Delta x= 7.5 m t= 1.626 s v= ? Delta x = vxt vx=...

Delta X = 5.0 m Vx= 4.613 m/s t =? Delta X = Vx(t) t= delta x/ vx = 5/4.613 = 1.083 s So do I add 1.083 to the previously establish 1.626 and then get 2.709 for total time?
14. ### Physics Puzzler!

HELP!!!!! I am in grade 12 physics and my teacher gave us a puzzler for the weekend. I need serious help with it and all of my classmates cant figure it out either! The question is : A 10.8 kg box is 5.0 m from the edge of a 12.96 m high cliff. Bob pushes the box over the edge of the cliff by...

Ok... so I got vx to be 4.613 m/s. Then I use Pythagoras to find the velocity (vy-squared- + vx-squared- = v-squared) and I got it to be 16.59. Then to find acceleration I went: a= V/t = 10.20 m/s-squared- Am I going right...because then I get F= ma F= 10.8 (10.20) =110.19 N I don't...

The equations in the last one are a little wack. sorry... © = squared ¥Äx= delta x

What I did... I will show you what I did and then can you please tell me what I'm doing wrong. Consider Vertical: ¥Äx= -12.96 m Vo= 0 m/s a= -9.80 m/s v= ? t =? v©÷ = vo©÷ + 2a¥Äx v©÷ = 0 + 2a¥Äx = 2a¥Äx = 2(-9.80)(-12.96) = 254.016 = ¡î254.016 =15.93 m/s v =...