Jump to content # RoyalXBlood

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## About RoyalXBlood

• Rank
Quark
• Birthday 07/28/1988

## Profile Information

• Location
USA, NC
• Interests
Guitar, Composing music, video games
• College Major/Degree
UNC at Charlotte - Phsyics Major
• Favorite Area of Science
Physics
• Biography
Love, music, and physics is my life.
• Occupation
Expeditor - Cracker Barrel
1. ## Problem velocity and banked curves. (Physics)

I'm assuming my difference in the final answer comes from miscalculating the correct normal force... how would I calculate it from the given information?
2. ## Problem velocity and banked curves. (Physics)

Would it be in the same direction as the 22422N? If so I wouldn't know what to do from there..
3. ## Problem velocity and banked curves. (Physics)

The problem: A concrete highway curve of 80 meters is banked at a 19 degree angle. What is the maximum speed with which a 1800kg rubber-tired car take this curve without sliding? (static coefficient of friction of rubber on concrete = 1) So would I set up something like this? The answer is 40.1m/s, but I get 31.6m/s. So I need help to figure out what I'm not getting. I have the force of gravity in the y direction equal to 16,679.0N and in the x direction it is equal to 5,743.0N. The max friction would equal the normal force since the coefficient is 1, so the max friction would equal
4. But isn't the limit of 1/$x^{.2}$ as x goes to zero equal to infinite?
5. The problem is the integral of 1/$x^{.2}$ from x=0 to x=8. Apparently it's not divergent, but the problem 1/$x^{1.4}$ is and I understand that. Why is one divergent and not the other, and how would I get the area of 1/$x^{.2}$ from x=0 to x=8?
6. ## Coefficient of kinetic friction...(physics)

Thank's for helping me out so much, I appreciate the dedication.
7. ## Coefficient of kinetic friction...(physics)

Ok say leg b goes along the yaxis and leg a goes along the x-axis. C is the hypotenuse. B=50m. The only thing that I am not certain of is the angel, the given angel is that the angel formed between b and c or a and c? Ok using 50/cos80 I get 287.29385242m, that must be it, then the force would be 138.9185421. The difference is 346.169065N. That would be $F_{k}$. So the coefficient would be .7173658791 or just .72. It's hard for me to understand because you basicly subtract the final force acting on the guy from the initial and that would get you the force of kenetic friction
8. ## Coefficient of kinetic friction...(physics)

I assume the equation I can use is $V^{2}$ = $V_{o}^{2}$ + 2a$\Delta$x. From that I get a=15.75692405 (I like the keep all the figures). So the force accerating him is 787.8462075 Newtons. I still don't see how I can use this to find Force of kentic friction.
9. ## Coefficient of kinetic friction...(physics)

Kieran takes off down a 50 m high, 10° slope on his jet-powered skis. The skis have a thrust of 400 N. The combined mass of skis and Kieran is 50 kg (the fuel mass is negligible). Kieran's speed at the bottom is 40 m/s. What is the coefficient of kinetic friction of his skis on snow? So far my work shows... I know that force due to mass and gravity is 490. I'll call it $F_{g}$. I found it's x and y components to be $F_{xg}$= 85.1 and $F_{yg}$= 482.6. So $F_{N}$ is 482.6. That gets me one step closer because I know I need to solve [math]F_{k}[
10. I guess I didn't really think about that because I was thinking about the distance the bolt would fly up and then down. The distance would be greater than what the rocket traveled but the formula only cares about displacement. And since the rocket went up and the bolt is going down I would slap a negative on one side to correctly evaluate it right? So would this be the right set up? -[0(4) + (1/2)a(4)^2] = 4a(7) + (1/2)(-9.8)(7)^2 If so, after a little algebra I get a=6.67m/s^2
11. 1) What are the forces on the rocket? It has a constant positive acceleration. 2) At the moment the bolt falls off the rocket, what are the forces on it? It will start to deaccelerate due to gravity. 3) What would be initial conditions of the bolt when it falls off the rocket? The initial velocity would be equal to the velocity of the rocket at the moment it fell off. So I know V(inital of bolt) = V(final of rocket) I know I can use V(final of rocket)=4a, which means the inital velocity of the bolt with be 4a. But I dont know where to go with that. And I know the bolt will
12. I really would like to figure out how to do this, it's the only problem I can't solve so far.. "A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 7.00 later." "What was the rocket's acceleration?" We are focusing on three equations: 1) v = v(initial) + at 2) (delta)x = vt + .5at^2 3) v^2 = v(initial)^2 + 2a(delta)x x = distance v = velocity a = acceleration t = time (delta)x = x(final) - x(initial)
13. I like you, Edtharan.
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