To summarize, an object moving in uniform circular motion is moving around the perimeter of the circle with a constant speed<span>. While the </span>speed<span> of the object is</span>constant<span>, its </span>velocity<span> is </span>changing<span>. </span>Velocity<span>, being a vector, has a </span>constant<span>magnitude </span>but<span> a </span>changing<span> direction.</span>
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:
According to studies, the milky way is approximately, "170,000–200,000 light-years (52–61 kpc) in diameter and, on average, approximately 1,000 ly (0.3 kpc) thick."
With that being said, it is safe to say that the dimensions are somewhere around 100,000 by 1,000
From the given information in the question, the correct option is Option 1: 14 cm.
A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:
PE =
k 
Where PE is the energy, k is the spring constant and x is extension.
i. Given that: PE = 10 J and x = 10 cm, then;
PE =
k 
10 =
k 
20 = 100k
k = 0.2 J/cm
ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.
PE =
k 
20 =
(0.2) 
40 = 0.2 
= 200
x = 
= 14.1421
x = 14.14 cm
So that;
x is approximately 14.00 cm.
Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.
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Answer:
The correct answer is 24.
Explanation: