Respuesta :
Answer:
Explanation:
Acceleration of particle A is 7.3 times the acceleration of particle B.
Let the acceleration of particle B is a, then the acceleration of particle A is
7.3 a.
Let the period of particle A is T and the period of particle B is 2.5 T.
Let the radius of particle A is RA and the radius of particle B is RB.
Use the formula for the centripetal force
[tex]a=r\omega ^{2}=r\times \frac{4\pi^{2}}{T^{2}}[/tex]
So, [tex]r = a\frac{T^{2}}{4\pi^{2}}[/tex]
The ratio of radius of A to the radius of B is given by
[tex]\frac{R_{A}}{R_{B}}=\frac{a_{A}\times T_{A}^{2}}{a_{B}\times T_{B}^{2}}[/tex]
[tex]\frac{R_{A}}{R_{B}}=\frac{7.3 a\times T^{2}}{a\times 6.25T^{2}}[/tex]
RA : RB = 1.17
Answer:
The ratio of he radius of the motion of particle A to that of particle B is closest to 1.16.
Explanation:
Let [tex]a_A,a_B\ and\ r_A,r_B[/tex] are accelerations of particle and radius of A and B respectively. It is given that :
[tex]a_A=7.3\times a_B\\\\\dfrac{a_A}{a_B}=7.3\\\\and\\\\T_B=2.5\times T_A\\\dfrac{T_B}{T_A}=2.5[/tex]
The centripetal acceleration is given by the formula as :
[tex]a=\dfrac{v^2}{r}[/tex]
r is the radius of motion
Since, [tex]v=\dfrac{2\pi r}{T}[/tex]
[tex]a=\dfrac{4\pi ^2r}{T^2}[/tex]
For A
[tex]a_A=\dfrac{4\pi^2 r_A}{T^2_A}[/tex].........(1)
For B
[tex]a_B=\dfrac{4\pi^2 r_B}{T^2_B}[/tex]...........(2)
Dividing equation (1) and (2) we get :
[tex]\dfrac{a_A}{a_B}=\dfrac{T^2_B\times r_A}{T^2_A\times r_B}[/tex]
[tex]\dfrac{r_A}{r_B}=\dfrac{a_A\times T^2_A}{a_B\times T^2_B}[/tex]
Now using given conditions :
[tex]\dfrac{r_A}{r_B}=\dfrac{7.3}{(2.5)^2}\\\\\dfrac{r_A}{r_B}=1.16[/tex]
So, the ratio of he radius of the motion of particle A to that of particle B is closest to 1.16.
