B. If you do 39-3/4 divided by 12-1/2 you get 3.18
Answer:
Normal force = 0.326N
Explanation:
Given that:
mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg
height of the mass = 1.1 m
radius = 0.2 m
acceleration due to gravity = 9.8 m/s²
We are to determine the normal force pressing on the track at A.
To to that;
Let consider the conservation of energy relation; which says:
mgh = mgr + 1/2 mv²
gh = gr + 1/2 v²
gh - gr = 1/2v²
g(h-r) = 1/2v²
v² = 2g(h-r)
However; the normal force will result to a centripetal force; as such, using the relation
N =mv²/r
replacing the value for v² = 2g(h-r) in the above relation; we have:
Normal force = 2mg(h-r)/r
Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2
Normal force = 0.065268/0.2
Normal force = 0.32634 N
Normal force = 0.326N
Answer:
Explanation:
The complete detailed explanation which answer the question efficiently is shown in the attached files below.
I hope it helps a lot !
Answer:
a. 0.9263
b. 0.4872
c. 13.83kN/m
Explanation:
moisture content (ω) = 0.32
void ratio (e) = 0.95
specific gravity (
) = 2.75
the degree of satruation (S) = 
=0.32×2.75/0.95 = 0.9263
b. porosity (n) = 
= 0.95/(0.95 + 1)= 0.4872
c. dry unit weight (γ
) = 
taking specific unit weight of water (V
)= 9.81kN/m
γ = 2.75 × 1000/(1 + 0.95) = 13.83kN/m
