transfer momentum

- anonymous

transfer momentum

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- schrodinger

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- anonymous

a car is travelling at a constant speed in a circle with radius 100 m

- anonymous

drawing is not quite clear, could you please draw it again? :P

- anonymous

|dw:1378457063394:dw| those are 2 degree

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## More answers

- anonymous

Okay, let's take a look... do we also have a measure for this angle:
The angle between r1 and r2?

- anonymous

ya it is 2 degree too... between r1 and r2, r2 and r3, r4 and r5, and r5 and r6 those are all 2 degree

- anonymous

Alright, thanks.

- anonymous

it looks like first we'll need to find the speed of the car going around the circle, by using the equation relating speed, time and distance:

- anonymous

|dw:1378457917828:dw|

- anonymous

100 m / 0.6

- anonymous

We were given the time to get from r1 to r3... and can also find the distance along the circle between those points, since we were given the circle's radius. But before we can get the distance between the r1 and r3, we'll need the distance around the whole circle (its circumference)
Do you remember the formula for circumference?

- anonymous

2 pi r

- anonymous

circumference = 628

- anonymous

its involved a little more than that, for the distance we need to do more calculations.

- anonymous

like...?

- anonymous

|dw:1378458211749:dw|

- anonymous

(628*2)/360 = distance between r1 and r2 = 3.49

- anonymous

then, we see that between each two points marked along the circle, that takes up exactly 1/12th of the whole circumference. Since the points are separated by equal angles...

- anonymous

hmm, are we sure the angles aren't marked as a symbol like: rather than 2 that would make a little more sense based on the picture

- anonymous

sorry but distance between 2 points is 4 degree (r1 to r3) which is 1/90 not 1/12

- anonymous

|dw:1378458471545:dw|

- anonymous

if it really is 2, then what you have is correct.

- anonymous

ya it is 2

- anonymous

okay, yes if those really are just 2 degrees each
I was interpreting it a little differently, but it's difficult to say without seeing the exact problem statement.

- anonymous

:P

- anonymous

|dw:1378458681252:dw|

- anonymous

3.49

- anonymous

thats the exact problem statement i wrote and the exact same figure

- anonymous

If those really are 2 degrees, then the distance between r1 and r3 would be given by 628*4/360, similar to what you wrote.

- anonymous

Now we can plug in this distance and time into the formula speed

- anonymous

yep can u go ahead and solve it please so if i have question i can ask later

- anonymous

So plugging in d= 3.49 and t=.6, we get s=3.49/.6
Could you tell me what that equals?

- anonymous

wait that is for 2 degree, for 4 degree it is 6.9

- anonymous

okay, thanks for checking

- anonymous

np...hey i havee real short time to solve that question like 30 minutes....can you do it asap

- anonymous

Alright, so that gives a speed around the circle 11.5 m/2 at point r2, the car is moving straight up since that's the point where it crosses the axis...

- anonymous

so what are the components

- anonymous

So at that point, the car's horizontal velocity is 0 and the vertical velocity is the full of 11.5 m/s

- anonymous

yep at r2 right

- anonymous

|dw:1378459259658:dw|

- anonymous

I've circled it on that board

- anonymous

''components'' just means the horizontal and vertical velocities written separately

- anonymous

and yes you're right! :)

- anonymous

Any other questions about this problem?

- anonymous

z componenet at r2?

- anonymous

0... since the car is travelling in a flat circle, it only has x and y components.

- anonymous

all done?

- anonymous

nope lol 3 more question if u dont mind

- anonymous

0.6 seconds pass between t_4 when car is at r_4 and t_6 when car is at r_6 determine x,y,z components of car's velocity vector when it is near r_5

- anonymous

Not right now, but maybe later. Good luck with your studies!

- anonymous

ty

- anonymous

I'm tired, sorry. lol :/

- anonymous

np

- anonymous

thx for explaining stuff

- anonymous

no worries, my pleasure. :P

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