Answer:
The force exerted is 6189.4 N
Solution:
As per the question:
Area of the book, A = [tex]0.22\times 0.28 = 0.0616\ m^{2}[/tex]
Mass of the book, m = 3 kg
Pressure of the book, P = [tex]1.0\times10^{5}\ N/m^{2} [/tex]
Now,
Pressure, P = [tex]\frac{F}{A}[/tex]
Force, F = PA = [tex]1.0\times10^{5}\times 0.0616 = 6160\ N[/tex]
Force on mass of 3 kg, F' = mg = [tex]3\times 9.8 = 29.4\ N[/tex]
Now, the force at the bottom is given by:
F'' = F + F' = 6160 + 29.4 = 6189.4 N
The force that the atmosphere exerts on the bottom of the book is about 6160 Newton
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The basic formula of pressure that needs to be recalled is:
Pressure = Force / Cross-sectional Area
or symbolized:
[tex]\large {\boxed {P = F \div A} }[/tex]
P = Pressure (Pa)
F = Force (N)
A = Cross-sectional Area (m²)
Let us now tackle the problem !
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Given:
Area of the book = A = 0.22 × 0.28 = 0.0616 m²
Mass of the book = m = 3 kg
Atmospheric Pressure = P = 1.0 × 10⁵ N/m²
Asked:
Force on the bottom of the book = F = ?
Solution:
[tex]P = F \div A[/tex]
[tex]F = P \times A[/tex]
[tex]F = 1.0 \times 10^5 \times 0.0616[/tex]
[tex]F = 6160 \texttt{ Newton}[/tex]
[tex]\texttt{ }[/tex]
The force that the atmosphere exerts on the bottom of the book is approximately equal to the force that the atmosphere exerts on the top of the book. Therefore these forces will be cancel out. If Anita holds the book , she only has to exert force equal to the weight of the book.
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Grade: High School
Subject: Physics
Chapter: Pressure
