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.
Here, you can calculate it's potential energy with respect to ground.
We know, U = mgh
Here, m = 75 Kg
g = 9.8 m/s² [ constant value for earth system ]
h = 300 m
Substitute their values into the expression:
U = 75 × 9.8 × 300
U = 220500 J
In short, Your Final Answer would be 220,500 J
Hope this helps!
Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N