1. Oct 3, 2011

naham1866

1. At a distance of 25km from the eye of a hurricane, the wind is moving at 180km/h in a circle. What is the magnitude of the centripetal acceleration, in meters per second squared, of the particles that make up the wind?

2. Relevant equations
Ac=V^2/r = 4∏^2r/T^2 = 4∏^2rf^2

3. The attempt at a solution
Ac=V^2/r

The answer is 0.10m/s but I just cant seem to get it right.

2. Oct 3, 2011

Staff: Mentor

You have the right equation in your attempt. Fill in the numbers for velocity V and radius r, and be sure to stay consistent in your units. Show us that work...

3. Oct 3, 2011

naham1866

Okay so I got Ac=180km/h^2/25km

Problem is that it doesn't give me 0.10m/s

4. Oct 3, 2011

Staff: Mentor

You need to fix the units. You need to convert everything into meters, kilograms and seconds (the mks standard SI system of units). If you mix units like seconds and hours up, you won't get the right answers.

So convert everything you are given into mks units, and plug those quantities into the equation. Carry units along in the equation, and if you have the same units in numerator and denominator, you can cancle them out. Like m/m = 1, and s^2/s = s, and so on.

Now show us what you get...

5. Oct 3, 2011

naham1866

Haha YES!!! I Did it!! I went out and on my way back home I remembered that I had to change from kilometers to meters, so here's what I did.

180km/h / 60 = 3m/min /60 = 0.05m/s
25km / 1000 = 0.025m
Ac = V^2/r
Ac = 0.05m/s^2/0.025m
Ac= .10m/s
YES! :D

6. Oct 4, 2011

naham1866

Thank you vey much for the help

7. Oct 5, 2011

Staff: Mentor

You are welcome. Learning to carry units along in your calculations is a huge trick, IMO. I still remember the first time I learned that wey back in undergrad many years ago.