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The focal length of a lens needed by a woman whose near point is 50cm from her eyes is 50cm.
To find the answer, we have to know about the focal length of correcting lens.
<h3>
How to find the focal length of
correcting lens?</h3>
- If x is the distance of nearest point of the defective eye and D is the least distance of distinct vision, then, the expression for focal length of the correcting lens will be,

- It is given that, the woman whose near point is 50cm from her eyes, assuming the least distance of distinct vision for a normal eye is 25cm. Thus, the focal length will be,

Thus, we can conclude that, the focal length of a lens needed by a woman whose near point is 50cm from her eyes is 50cm.
Learn more about the focal length here:
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Answer:

Explanation:
The cabinet does not move: this means that the net force acting on it is zero.
Along the horizontal direction, we have two forces:
- The push exerted by Bob, F = 200 N, forward
- The frictional force,
, which acts in the opposite direction (backward)
Since the net force must be zero, we have:

So solving the equation we can find the magnitude of the friction force:


<u>Explanation:</u>
Velocity of B₁ = 4.3m/s
Velocity of B₂ = -4.3m/s
For perfectly elastic collision:, momentum is conserved

where,
m₁ = mass of Ball 1
m₂ = mass of Ball 2
v₁ = initial velocity of Ball 1
v₂ = initial velocity of ball 2
v'₁ = final velocity of ball 1
v'₂ = final velocity of ball 2
The final velocity of the balls after head on elastic collision would be

Substituting the velocities in the equation

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.
F(of spring)=230x=ma=3.5(5)=17.5=230x; x=0.07m.