Answer:
Energy. They need energy.
Explanation:
Answer:
The range of powers is 
Explanation:
From the question we are told that
The far point of the left eye is 
The near point of the left eye is 
The near point with the glasses on is 
From these parameter we can see that with the glass on that for near point the
Object distance would be 
Image distance would be 
To obtain the focal length we would apply the lens formula which is mathematically represented as

substituting values


converting to meters


Generally the power of the lens is mathematically represented as

Substituting values


From these parameter we can see that with the glass on that for far point the
Object distance would be 
Image distance would be 
To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

substituting values


converting to meters

Generally the power of the lens is mathematically represented as

Substituting values


This implies that the range of powers of the lens in his glass is

Answer: 
The following vectors have been given: 
The angle between these two vectors can be found by:




If a man pushes on a wall with some force then according to Newton's third law, wall will also apply force on man with same magnitude but opposite in direction.
Answer:When white light strikes an object, each individual frequency of light is transmitted, reflected, or absorbed, depending on the properties of the surface molecules. If all frequencies are absorbed by the object, then it appears black. If all frequencies are reflected, then it appears white.
Explanation: