We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the <em>book is</em>
From the question we are told
A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal
Generally the equation for the Force is mathematically given as

F=44N
Therefore
the minimum force that must be applied on the <em>book is</em>
F=44N
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Answer:
Total displacement will be 47 meter
Total distance will be 83 meters
Explanation:
We have given that first the student go eastward towards bus stop 20 meters
But he realizes that she dropped his physics notebook and so h=she turns back along the same way up to 18 meters
So displacement = 20-18 = 2 meters
And he travel 45 meters in east along the bus stop so total displacement = 45+2 = 47 meters
Total distance traveled by the student = 20+18+45 = 83 meters
The final mass after decay can be obtained by using under given relation:
half life period of As-81 = 33 seconds
mf = mi x (1/2^n)
= 100 x ( 1/2^(43.2/33))
= 40.4 %
Answer:
a)
, b) 
Explanation:
a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

The internal energy for a monoatomic ideal gas is:

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:



The thermal energy contained by the gas is:


b) The physical model for the cat is constructed from Work-Energy Theorem:

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:


