Answer:

Explanation:
Firstly finding 
So,

Substitute 

Exchange the values of x and y

Solving for y

Adding 5 to both sides

Dividing both sides by 10

Replace 

For x = -4


Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
- Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.
- It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:

- Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.
∴ ωp = ωw (2)
⇒ 

- Dividing (4) by (3), from (2), we have:


Imagine a skinny straw in the water, standing right over the hole. The WEIGHT of the water in that straw is the force on the tape. Now, the volume of water in the straw is (1 mm^2) times (20 cm). Once you have the volume, you can use the density and gravity to find the weight. And THAT's the force on the tape. If the tape can't hold that force, then it peels off and the water runs out through the hole. /// This is a pretty hard problem, because it involved mm^2, cm, and m^3. You have to be very very very careful with your units as you work through this one. If you've been struggling with it, I'm almost sure the problem is the units.
A peak = A Rms x Sq root 2
Therefore 3.6 x sq root of 2
A peak = 5.09
Using the given equation:
di = 20.0 * 10.0 / 20.0 - 10.0
di = 200/10
di = 20.0 cm
The answer is A.