anonymous
  • anonymous
Two students start at the northernmost point of the pool and walk slowly around it in opposite directions.(a)If the angular speed of the student walking in the clockwise direction (as viewed from above) is 0.050 rad/s and the angular speed of the other student is 0.015 rad/s, how long does it take before they meet? By using (0.015)t+(0.05)t=2pi I got t to equal 96.7 seconds (b) At what angle, measured clockwise from due north, do the students meet? (c) If the difference in linear speed between the students is 0.23 m/s, what is the radius of the fountain?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
For c I tried using the equation v=2pir/T using 96.7 seconds but realized thats not the full period to get around the fountain. I need to know b but I have no idea how to get the angle in the problem. Any help is appreciated
fretje
  • fretje
|dw:1433679913436:dw|
fretje
  • fretje
from use of following relation \[\alpha = \omega * t\] angle alpha = angular velocity * time

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

fretje
  • fretje
for student 1, S1 i called him/her: 0.05rad/s * 96.7s = 4.835 rad that is in degrees 4.835 rad * 360°/(2pi) = 277° for answer c, i can use the relation for tangential speed and angular velocity \[v _{\tan} = \omega * radius\] we know omega 1 and omega 2 from student S1 and S2
fretje
  • fretje
so with both equations and for both the same radius i get when i divide the one by the other: \[\frac{ v _{1} }{ v _{2} } = \frac{ \omega _{1} }{ \omega _{2} }\] i also have \[v_{1} - v _{2} = 0.23 m/s\] i can solve both equations for v2 and i get \[v _{2} = 0.09857 m/s\] so v1 \[v _{1} = 0.23 m/s +v _{2} = 0.3285 m/s\]
fretje
  • fretje
in previous i didnt mention this: \[\omega _{1}/\omega _{2} = 3.3333333\] but we knew that. Then from \[r = \frac{ v _{2} }{ \omega _{2} } = 0.09857 / 0.015 = 6.57 m\]
anonymous
  • anonymous
Thank you so much!!!!!
fretje
  • fretje
i hope it is correct.
anonymous
  • anonymous
it was! Thaank you!

Looking for something else?

Not the answer you are looking for? Search for more explanations.