Answer:
Radius of the circle will be 2.5 m
Explanation:
We have given velocity of particle moving in the circle v = 5 m/sec
Acceleration of particle in the circle ![a=10m/sec^2](https://tex.z-dn.net/?f=a%3D10m%2Fsec%5E2)
We have to find the radius of the circle
We know that acceleration is given by ![a=\frac{v^2}{r}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2%7D%7Br%7D)
So ![10=\frac{5^2}{r}](https://tex.z-dn.net/?f=10%3D%5Cfrac%7B5%5E2%7D%7Br%7D)
![r=\frac{25}{10}=2.5m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B25%7D%7B10%7D%3D2.5m)
So radius of the circle will be 2.5 m
a) 1.57 m/s
The sock spins once every 2.0 seconds, so its period is
T = 2.0 s
Therefore, the angular velocity of the sock is
![\omega=\frac{2\pi}{T}=\frac{2\pi}{2.0}=3.14 rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D%3D%5Cfrac%7B2%5Cpi%7D%7B2.0%7D%3D3.14%20rad%2Fs)
The linear speed of the sock is given by
![v=\omega r](https://tex.z-dn.net/?f=v%3D%5Comega%20r)
where
is the angular velocity
r = 0.50 m is the radius of the circular path of the sock
Substituting, we find:
![v=(3.14)(0.50)=1.57 m/s](https://tex.z-dn.net/?f=v%3D%283.14%29%280.50%29%3D1.57%20m%2Fs)
B) Faster
In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:
![r' = 2r = 1.00 m](https://tex.z-dn.net/?f=r%27%20%3D%202r%20%3D%201.00%20m)
At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:
![\omega' = \omega = 3.14 rad/s](https://tex.z-dn.net/?f=%5Comega%27%20%3D%20%5Comega%20%3D%203.14%20rad%2Fs)
Therefore, the new linear speed would be:
![v'=\omega' r' = \omega (2r)](https://tex.z-dn.net/?f=v%27%3D%5Comega%27%20r%27%20%3D%20%5Comega%20%282r%29)
And substituting,
![v'=(3.14)(1.00)=3.14 rad/s = 2v](https://tex.z-dn.net/?f=v%27%3D%283.14%29%281.00%29%3D3.14%20rad%2Fs%20%3D%202v)
So, we see that the linear speed has doubled.
Answer: Object B
Explanation: Acceleration is directly proportional to force and inversely proportional to mass. It implies that more massive objects accelerates at a slower rate.
J can get answer on this way:
Ek=m*V*V/2= (24kg*2m/s*2m/s)/2=48 Ј