Answer:
The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.
Explanation:
The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.
It is composed of the following six parts:
1. Skull (22 bones)
2. Ossicles of the middle ear
3. Hyoid bone
4. Rib cage
5. Sternum
6. Vertebral column
The axial region of the body forms the vertical axis of the body as the axial skeleton supports the head, neck, back, and chest.
Answer:
You pull on the oars. By the third law, the oars push back on your hands, but that’s irrelevant to the motion of the boat. The other end of each oar (the blade) pushes against the water. By the third law, the water pushes back on the oars, pushing the boat forward.
<span>Her center of mass will rise 3.7 meters.
First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so
8.5 m / 9.8 m/s^2 = 0.867346939 s
And the distance a object under constant acceleration travels is
d = 0.5 A T^2
Substituting known values, gives
d = 0.5 9.8 m/s^2 (0.867346939 s)^2
d = 4.9 m/s^2 * 0.752290712 s^2
d = 3.68622449 m
Rounded to 2 significant figures gives 3.7 meters.
Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.</span>
Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m
