Answer:
single replacement reaction
Fibrous joint functions as a suture to tightly bind bones together so they do not move.
We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:
Vf² = Vi² + 2ad
Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.
Given values:
Vi = 0m/s (dumbbell starts falling from rest)
a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)
d = 80×10⁻²m
Plug in the values and solve for Vf:
Vf² = 2(10)(80×10⁻²)
Vf = ±4m/s
Reject the negative root.
Vf = 4m/s
The momentum of the dumbbell is given by:
p = mv
p is its momentum, m is its mass, and v is its velocity.
Given values:
m = 10kg
v = 4m/s (from previous calculation)
Plug in the values and solve for p:
p = 10(4)
p = 40kg×m/s
Answer:
the standard way the body is positioned when using anatomical terminology ... invisible line that runs vertically through the center of the axial region.
Explanation:
