Answer:
The value  is   
 
Explanation:
From the question we are that 
      The velocity  
       The  time taken is   
 
      The  total mass of the train is  
 Generally the average power delivered is mathematically represented as 
       
    
       
    
=>   
    
=>   
   
 
        
                    
             
        
        
        
Answer:
5 N
Explanation:
The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N 
 
        
                    
             
        
        
        
Answer:

Explanation:
To solve this problem we use the formula for accelerated motion:

We will take the initial position as our reference ( ) and the downward direction as positive. Since the rock departs from rest we have:
) and the downward direction as positive. Since the rock departs from rest we have:

Which means our acceleration would be:

Using our values:

 
        
             
        
        
        
Answer:
The height is  
A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter ) 
Explanation:
    From the question we are told that 
            The height is 
             The angle of the slope is 
According to the law of conservation of energy 
      The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy 
                           
Where I is the moment of inertia which is mathematically represented as this for  a sphere 
                     
   The angular velocity  is mathematically represented as
 is mathematically represented as 
                          
So the equation for conservation of energy becomes 
                ![mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5B%5Cfrac%7B2%7D%7B5%7D%20mr%5E2%20%5D%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
               ![mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B2%7D%7B5%7D%20%2B1%20%5D)
              ![mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B7%7D%7B5%7D%20%5D)
             ![gh_s =[\frac{7}{10} ] v^2](https://tex.z-dn.net/?f=gh_s%20%3D%5B%5Cfrac%7B7%7D%7B10%7D%20%5D%20v%5E2)
               
Considering a circular hoop 
    The moment of inertial is different for circle and it is mathematically represented as 
              
Substituting this into the conservation equation above 
               ![mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2](https://tex.z-dn.net/?f=mgh_c%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28mr%5E2%29%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%20%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Where  is the height where the circular hoop would be released to equal the speed of the sphere at the bottom
 is the height where the circular hoop would be released to equal the speed of the sphere at the bottom
                  
                      
                      
 Recall that   
                     
                       
             Substituting values 
                    
                         
     
         
       
                          
  
        
             
        
        
        
Answer:
Explanation:
electric field at the location of electron 
= 9 x 10⁹ x 7.2 / .03²
= 72 x 10¹² N/C 
force on electron = electric field x charge on electron 
= 72 x 10¹² x 1.6 x 10⁻¹⁹
= 115.2 x 10⁻⁷ N . 
C ) 
work done = charge on electron x potential difference at two points 
potential at .03 m 
= 9 x 10⁹ x 7.2 / .03 
= 2.16 x 10¹² V 
potential at .001 m 
= 9 x 10⁹ x 7.2 / .001 
= 64.8 x 10¹² V 
potential difference = (64.8 - 2.16 )x 10¹² V 
= 62.64 x 10¹² V  . 
work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹
= 100.224 x 10⁻⁷ J . 
D ) 
There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction . 
Work done in case of electron will be positive and work done in case of positron will be negative . 
electric field due to charge will be same in both the cases .