Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;

Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.


f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;

Here, p is the power of the lens.
Put f= 35.71 cm.

p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
<span>Rising or falling, it does not change.</span>
Answer:
Explanation:
Given
mass of wheel m=13 kg
radius of wheel=1.8 m
N=469 rev/min

t=16 s
Angular deceleration in 16 s


Moment of Inertia 
Change in kinetic energy =Work done
Change in kinetic Energy

(a)Work done =50.79 kJ
(b)Average Power

Answer:
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Answer:
The idea behind space expansion is that after the big bang, the universe started infinitely expanding. It's also called metric expansion. Stephen Hawking's A Brief History of Time goes further into detail on this.