Recent content by Jess048

  1. J

    How Does a Rolling Ball's Velocity Change with Constant Acceleration?

    Sry, but this still leaves me with the same problem I already figured out question a. I know the formula for question b, but my problem is I don't know what to put for displacement. How would i figure out d1 and d2 in the formula. For question c I am not sure how to solve this question so how...
  2. J

    How Does a Rolling Ball's Velocity Change with Constant Acceleration?

    Question: A rolling ball has an initial velocity of -1.5 m/s. a. If the ball has a constant acceleration of -0.23 m/s^2, what is its velocity after 2.2s? b. What was its average velocity during that time interval? c. How far did it travel in this time interval? So far i got: a. V=...
  3. J

    Understanding Trigonometric Functions: Period and Phase Shift Explained

    The period of function y = tan k0 is pie/k where k>o. I'm not sure about the phase shift.
  4. J

    Newton's Variation of Kepler's third law

    Calculate the period of the Earth's moon if the radius of orbit was twice the actual value of 3.9 x 10^8 m. a. 1.13 x 10^6 s or 13 days b. 2.3 x 10^6 s or 26 days c. 5.14 x 10^6 s or 59 days d. 6.85 x10^6 s or 79 days So far I got: v=velocity; T=time; h= hieght from surface v=? T=...
  5. J

    Understanding Trigonometric Functions: Period and Phase Shift Explained

    State the period and phase shift of the function y= -4 tan (1/2x + 3pie/8). I know the answer is pie; -3pie/8, but I don't understand the process could someone explain how the period and phase shift are found in these functions.
  6. J

    Question: Kepler's third law of planetary motion

    The mean distance between the Earth and the moon is 3.84x10^8 m, and the moon has an orbital period of 27.3 days. Find the distance from Earth of an artificial satellite that has an orbital period of 8.5 days. a. 1.76x10^8 m b. 1.76x10^4 m c. 1.76x10^10 m d. 5.24x10^8 m So far I got: I...
  7. J

    What is the parallel-component of the weight

    Oh ok so the answer would be C
  8. J

    What is the parallel-component of the weight

    A 50 kg trunk rest on a ramp at 18 degrees. What is the parallel-component of the weight? a. 15.5 N b. 47.6 N c. 151 N d. 466 N So far I got: I used the formula Fgx = W sin 0. Fgx= -(490 N) sin(18) My answer was 368, as you see it is not one of the choices. Am i leaving out a step...
  9. J

    Motion of a jumper - Find the mass

    A high jumper falling at 3.9 m/s, lands on a foam pit and comes to rest, compressing the pit a distance of 0.43 m. If the pit is able to exert an average force of -1100 N on the high jumper in breaking the fall, what is the jumper's mass? I don't know where to begin so anything can...
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