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:
Differences
microscopic refers to substances visible to the naked eye
macroscopic are substances invisible to naked eye
Similarities
both refer to different scales that are useful to determining the size to different compounds.
Explanation:
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We must remember that the total net force equation at
constant velocity is:
<span>F – Ff = 0</span>
of
F - µN = 0
Using Newton's 2nd Law of Motion:<span>
F = m a
<span>Where,
F = net force acting on the body
m = mass of the body
a = acceleration of the body
Since the cart is moving at a constant velocity, then
acceleration is zero, hence the working equation simplifies to
F = net Force = 0
Therefore,
F - µN = 0
where
µ = coefficient of friction = 0.20
N = normal force acting on the cart = 12 N
Therefore,
F - 0.20(12) = 0
<span>
F = 2.4 N </span></span></span>
Answer:
Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd
Explanation:
The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is
d = 12 + 9 = 21 yd
The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:

Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was
d = 15 - 3 = 12 yd