Answer:
2.74
Explanation:
Magnification = image distance/object distance
Mag = v/u
Given
v = 46cm
u = 16.8
Magnification = 46/16.8
Magnification = 2.74
Hence the magnification is 2.74
Answer:

Explanation:
We are asked to find the force being applied to a book. According to Newton's Second Law of Motion, force is the product of mass and acceleration.

The mass of the book is 0.75 kilograms and the acceleration is 0.3 meters per square second. Substitute these values into the formula.

Multiply.

1 kilogram meter per second squared is equal to 1 Newton. Therefore, our answer of 0.225 kilogram meters per second squared is equal to 0.225 Newtons.

<u>0.225 Newtons of force</u> are applied to the book.
Answer:
0.03167 m
1.52 m
Explanation:
x = Compression of net
h = Height of jump
g = Acceleration due to gravity = 9.81 m/s²
The potential energy and the kinetic energy of the system is conserved

The spring constant of the net is 20130.76 N
From Hooke's Law

The net would strech 0.03167 m
If h = 35 m
From energy conservation

Solving the above equation we get

The compression of the net is 1.52 m