Search results

  1. P

    Murder scene

    So I just solve dy/dt= -.1 when t=0 and that should give me my k?
  2. P

    Murder scene

    1 the problem and all known variables. A person is murdered in a room with a temperature of 20 deg C. At the time the body is discovered, the body temp is 32 deg C and is decreasing at an instantaneous rate of .1 deg C/minute. How long ago was the murder commited? 2. Related equations...
  3. P

    Pain measurement

    I was just wondering if there's a way to measure pain. I figured it'd be and equation like current. Like voltage would be the amount of pain, ohm would be a persons resistance to pain, and current would be the pain they actually felt.
  4. P

    Einstein and the graviton

    We know that there are force carrying particles for the strong and weak nuclear force, and the electromagnetism force. But in Einsteins theory of relativity, he states that gravity is the bending of space time, not a force. So if there is no force, why do we say there must be a graviton? Ps I'm...
  5. P

    Spring motion

    Ok i fixed the two mistakes, but I'm still coming up with the wrong answer. I think I have something wrong with the peiod but am unsure how to fix it
  6. P

    Spring motion

    Homework Statement When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled down from this position and released. It undergoes 57 oscillations in 39.0 s. What was the distance d? Homework Equations F=-Kx F=mg T=2pi*sqrt(K/m) The...
  7. P

    Asteroid Gravity

    Thank you. The answer was looking for you to say that 113km was the distance between the center of the asteroid.
  8. P

    Asteroid Gravity

    @ D H I see... I don't know why I have the sqrt in the g equation... But even so, I'm getting .01127 m/s^2 and its still wrong. @rude man I haven't used the same R.. I used 1323000 m for the top R and 19300 for the bottom R
  9. P

    Asteroid Gravity

    Ok so for units I'm getting sqrt(m)/s... which I have no idea what that means...
  10. P

    Two Spheres coming together

    For r1 I have .121+.353+.121 which is .595 m for r2, it's just .242 m
  11. P

    Asteroid Gravity

    That's 19.3 km in m. So I get g= sqrt((4π^2*(132300)^3)/(19300^2 * 147600^2)) (132300m is 19.3km+113km.) Which comes to .106 m/s^2, but it still says its the wrong answer.
  12. P

    Two Spheres coming together

    Homework Statement Two identical 28.5-kg spheres of radius 12.1 cm are 35.3 cm apart (center-to-center distance) and at rest in outer space. (You can assume that the only force acting on each mass is the gravitational force due to the other mass.) a) If they are released from rest and...
  13. P

    Asteroid Gravity

    I have (4 π^2 (19300m+113000m)^3)/((6.674*10^(-11)) (147600s)) which came out to be 6.28753*10^(16) kg So then I have g=sqrt(G M/R^2) or g=sqrt((6.674*10^(-11)) (6.28753*10^(16)))/ (19300^2) Which came out to be 14.74 m/s^2 (sorry about the exponents being...
  14. P

    Asteroid Gravity

    Ok. So I put the numbers in and got a mass of 6.2875.... E16. So then using the second equation I came up with 14.74 m/s^2 which is totally unreasonable for and asteroid that small. I changed all the kilometers to meters so the units all go together, so I'm not sure where I went wrong.
  15. P

    Asteroid Gravity

    Homework Statement An asteroid is discovered to have a tiny moon that orbits it in a circular path at a distance of 113 km and with a period of 41.0 h. The asteroid is roughly spherical (unusual for such a small body) with a radius of 19.3 km. a) Find the acceleration of gravity at the...
  16. P

    Orbital Period

    Thanks both of you. @ Cephid, you were right. I must of typed numbers wrong the first time.
  17. P

    Orbital Period

    Homework Statement A planet with a mass of 8.99·1021 kg is in a circular orbit around a star with a mass of 1.33·1030 kg. The planet has an orbital radius of 1.21·1010 m. a) What is the linear orbital velocity of the planet? b) What is the period of the planets orbit? c) What is the...
  18. P

    Rocket propulsion

    I found the final velocity, but when i tried to use the formula for distance, I found that I didn't have r sub p or T max and have no clue how to find them.
  19. P

    Rocket propulsion

    Homework Statement A rocket in outer space has a payload of 4050.0 kg and 1.753·105 kg of fuel. The rocket can expel propellant at a speed of 4.300 km/s. Assume that the rocket starts from rest, accelerates to its final velocity, and then begins its trip. How long will it take the rocket to...
  20. P

    Puck on Ice

    Then how do I find acceleration? I don't have time..
  21. P

    Puck on Ice

    A hockey puck on a frozen pond with an initial speed of 12.3 m/s stops after sliding a distance of 198.9 m. Calculate the average value of the coefficient of kinetic friction between the puck and the ice. So I started with finding the acceleration. I divided 12.3m/s by 198.9m then took the...
  22. P

    Motion of a Soccer Ball

    ohh.. I have bad handwritind -_- thank you. I did get the right answers
  23. P

    Motion of a Soccer Ball

    sorry* guess I still am -_-
  24. P

    Motion of a Soccer Ball

    dory, I was braindead last night. I figured out that the max height is 10.82 m and the range is 13.75 m. although the last one is still bothering me. I found the time it took the ball to hit the ground from height=10.82 which was about 1.47 s. I used 1.47 in Vy=gt (because Vy at max height is...
  25. P

    Motion of a Soccer Ball

    Neglect air resistance for the following. A soccer ball is kicked from the ground into the air. When the ball is at a height of 10.07 m, its velocity is (vx,vy) = (4.63,3.83) m/s. To what maximum height will the ball rise? What horizontal distance will be traveled by the ball? With what...
  26. P

    Postion problem

    ok thank you for your help!
  27. P

    Postion problem

    It comes to 15 when you put 25/6 which is 4.16 repeating. But yes the site does care about significant figures.
  28. P

    Postion problem

    We are given a velocity vs. time graph. It starts at 15 m/s at t=0 and goes to -10m/s at t=4 then it goes from -10m/s at t=4 to 15m/s at t=10 and we are told to find the postion after t=10 I first found the slope of the first segment which was -6.25. so I did ∫-6.25t+15 dt from 0 to 4 and...
  29. P

    Stuck with an acceleration problem

    Oh jeez.. Thank you. This is just a step in a bigger problem and I forgot how to do this part. Thank you rockfreak!
  30. P

    Stuck with an acceleration problem

    I know this is quite simple question. An object has an into all velocity of 50m/s it's traveling straight up. Assume g=-10m/s^2 how long does it take to go up and down? I put these numbers into x=1/2at^2+vi*t but I still have 2 variables. X and T. Idk what to do.
Top