Answer:
<h3>please type full question. Does not mean what you are trying to saying. </h3>
Answer:
μ = 0.18
Explanation:
Let's use Newton's second Law, the coordinate system is horizontal and vertical
Before starting to move the box
Y axis
N-W = 0
N = W = mg
X axis
F -fr = 0
F = fr
The friction force has the formula
fr = μ N
fr = μ m g
At the limit point just before starting the movement
F = μ m g
μ = F / m g
calculate
μ = 34.8 / (19.8 9.8)
μ = 0.18
Answer:
B. w=12.68rad/s
C. α=3.52rad/s^2
Explanation:
B)
We can solve this problem by taking into account that (as in the uniformly accelerated motion)
( 1 )
where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.
In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have


to calculate the angular speed w we can use
Thus, wf=12.68rad/s
C) We can use our result in B)

I hope this is useful for you
regards
Answer:

Explanation:
R = Horizontal range of projectile = 75 m
v = Velocity of projectile = 37 m/s
g = Acceleration due to gravity = 
Horizontal range is given by

The angle at which the arrow is to be released is
.
I think that the answer is friction