The answer is “ The medium”
When you kick a ball the when your foot hits the ball the force from your foot is now on the ball and if the ball bumps into somoene they might fall(if you kick hard)
<span>B. Reviewing the competency of health care workers</span> is considered part of health care operations
<h2>
Component of the velocity of the ball in the horizontal direction just before the ball hits the ground = 7.31 m/s</h2>
Explanation:
In horizontal direction there is acceleration or deceleration for a ball tossed upward at an initial angle of 43° off the ground.
So the horizontal component of velocity always remains the same.
Horizontal component of velocity is the cosine component of velocity.
Initial velocity, u = 10 m/s
Angle, θ = 43°
Horizontal component of velocity = u cosθ
Horizontal component of velocity = 10 cos43
Horizontal component of velocity = 7.31 m/s
Since the horizontal velocity is unaffected, we have
Component of the velocity of the ball in the horizontal direction just before the ball hits the ground = 7.31 m/s
Since the ladder is standing, we know that the coefficient
of friction is at least something. This [gotta be at least this] friction
coefficient can be calculated. As the man begins to climb the ladder, the
friction can even be less than the free-standing friction coefficient. However,
as the man climbs the ladder, more and more friction is required. Since he
eventually slips, we know that friction is less than what's required at the top
of the ladder.
The only "answer" to this problem is putting lower
and upper bounds on the coefficient. For the lower one, find how much friction
the ladder needs to stand by itself. For the most that friction could be, find
what friction is when the man reaches the top of the ladder.
Ff = uN1
Fx = 0 = Ff + N2
Fy = 0 = N1 – 400 – 864
N1 = 1264 N
Torque balance
T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)
N2 = 439 N
Ff = 439= u N1
U = 440 / 1264 = 0.3481
