Explanation:
Given that,
Mass, m = 0.08 kg
Radius of the path, r = 2.7 cm = 0.027 m
The linear acceleration of a yo-yo, a = 5.7 m/s²
We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
(a) Tension :
The net force acting on the string is :
ma=mg-T
T=m(g-a)
Putting all the values,
T = 0.08(9.8-5.7)
= 0.328 N
(b) Angular acceleration,
The relation between the angular and linear acceleration is given by :

(c) Moment of inertia :
The net torque acting on it is,  , I is the moment of inertia
, I is the moment of inertia
Also, 
So,
 
Hence, this is the required solution.
 
        
             
        
        
        
Answer:

Explanation:
 = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C
 = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C 
 = mass of each proton = 1.67 x 10⁻²⁷ kg
 = mass of each proton = 1.67 x 10⁻²⁷ kg 
r = distance between the two protons = 1 x 10⁻¹⁵ m 
Electric force between the two protons is given as 


 N
 N 
Gravitational force between the two protons is given as 


 N
 N 
Ratio is given as 



 
        
                    
             
        
        
        
It decreases and the decrease in the energy level appear as a kinetic energy of the charge