Answer: 
a= 0.5m/s^2
Explanation:
Force applied on an object is known as 
F=m.a  (Newton's second law states it)
a=F/m 
a=5/10=0.5m/s^2
 
        
             
        
        
        
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is 
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

where
 is the vacuum permeability
 is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:

 
        
             
        
        
        
Answer:
206.8965517 n
Explanation:
First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.
 
        
             
        
        
        
You can use Vf^2-Vi^2 = 2ax
Vf^2 - 0 = 2(9.81)(25)
Or you can use energy
mgh = 1/2mv^2
2gh =v^2
Same thing