It's called cellular differentiation. I think.
Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
The answer is d I took the test
Answer:
The x-component of 
is 56.148 newtons.
Explanation:
From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:
(1)
Where:
, 
, 
- External forces exerted on the ring, measured in newtons.
- Vector zero, measured in newtons.
If we know that ![\vec F_{1} = (70.711,70.711)\,[N]](https://tex.z-dn.net/?f=%5Cvec%20F_%7B1%7D%20%3D%20%2870.711%2C70.711%29%5C%2C%5BN%5D)
, ![\vec F_{2} = (-126.859, 46.173)\,[N]](https://tex.z-dn.net/?f=%5Cvec%20F_%7B2%7D%20%3D%20%28-126.859%2C%2046.173%29%5C%2C%5BN%5D)
, 
and ![\vec O = (0,0)\,[N]](https://tex.z-dn.net/?f=%5Cvec%20O%20%3D%20%280%2C0%29%5C%2C%5BN%5D)
, then we construct the following system of linear equations:
(2)
(3)
The solution of this system is:
, 
The x-component of is 56.148 newtons.
Answer:
when you open a can of pop
when you jump on your bed
Explanation:
