Answer:
Abdominal
Sitting up, postural alignment
Biceps
Lifting, pulling
Deltoids
Overhead lifting
Erector Spinae
Postural alignment
Gastronemius & Soleus
Push off for walking, standing on tiptoes
Gluteus
Climbing stairs, walking, standing up
Hamstrings
Walking
Latissimus Dorsi & Rhomboids
Postural alignment, pulling open a door
Obliques
Rotation and side flexion of body
Pectoralis
Push up, pull up, bench press
Quadriceps
Climbing stairs, walking, standing up
Trapezius
Moves head sideways
Triceps
Pushing
God bless you. Because my soul almost left my body when i had to do this.
Answer:
The final acceleration becomes (1/3) of the initial acceleration.
Explanation:
The second law of motion gives the relationship between the net force, mass and the acceleration of an object. It is given by :

m = mass
a = acceleration
According to given condition, if the mass of a sliding block is tripled while a constant net force is applied. We need to find how much does the acceleration decrease.

Let a' is the final acceleration,

m' = 3m



So, the final acceleration becomes (1/3) of the initial acceleration. Hence, this is the required solution.
Answer:
4
Explanation:
We are given that

K.E at x=0 m=20 J
K.E at x=3 m=11 J
We have to find the value of c.
By work energy theorem
Work done=Change in kinetic energy
W=
![W=[\frac{cx^2}{2}-x^3]^{3}_{0}](https://tex.z-dn.net/?f=W%3D%5B%5Cfrac%7Bcx%5E2%7D%7B2%7D-x%5E3%5D%5E%7B3%7D_%7B0%7D)






Answer:
1.2 seconds
Explanation:
distance = ((final speed + initial speed) * time)/2
Here given:
Solving steps:
3.8 = ((0 + 6.4) * time))/2
3.8 = 3.2(time)
time = 3.8/3.2
time = 1.1875 seconds ≈ 1.2 seconds
Answer:
You're four sentences should include about how the roller coaster has the most potential energy at the top of the track, and the opposing energy, "kinetic" has the most kinetic energy when going down the hill.
Explanation:
Kinetic - In-Motion.
Potential - Gathering Energy to go into Motion.
( I'll try to answer questions to clear up confusion. )