Answer:
0.023 Pa*s
Explanation:
The surface area of the side of the inner cylinder is:
A = π*d*l
A = π*0.15*0.75 = 0.35 m^2
At 200 rpm the inner cylinder has a tangential speed of:
u = w * r
u = w * d/2
w = 200 rpm * 2π / 60 = 20.9 rad/s
u = 20.9 * 0.15 / 2 = 1.57 m/s
The torque is of 0.8 N*m, this means that the force is:
T = F * r
F = T / r
F = 2*T / d
For Newtoninan fluids with two plates moving respect of each other with a fluid between the viscous friction force would be:
F = μ*A*u / y
Where
μ: viscocity
y: separation between pates
A: surface area of the plates
Then:
2*T / d = μ*A*u/y
Rearranging:
μ = 2*T*y / (d*A*u)
μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s
Answer:
The distance measure from the wall = 36ft
Explanation:
Given Data:
w = 10
g =32.2ft/s²
x = 2
Using the principle of work and energy,
T₁ +∑U₁-₂ = T₂
0 + 1/2kx² -wh = 1/2 w/g V²
Substituting, we have
0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²
170 = 0.15528V²
V² = 170/0.15528
V² = 1094.796
V = √1094.796
V = 33.09 ft/s
But tan ∅ = 3/4
∅ = tan⁻¹3/4
= 36.87°
From uniform acceleration,
S = S₀ + ut + 1/2gt²
It can be written as
S = S₀ + Vsin∅*t + 1/2gt²
Substituting, we have
0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)
19.85t - 16.1t² + 3 = 0
16.1t² - 19.85t - 3 = 0
Solving it quadratically, we obtain t = 1.36s
The distance measure from the wall is given by the formula
d = VCos∅*t
Substituting, we have
d = 33.09 * cos 36. 87 * 1.36
d = 36ft
Answer:
hello your question lacks the required question attached below is the missing diagram
Forces in GJ = -4.4444 i.e. 4.4444 tons
Forces in IG = 15.382 tons ( T )
Explanation:
Forces in GJ = -4.4444 i.e. 4.4444 tons
Forces in IG = 15.382 tons ( T )
attached below is the detailed solution