Which of the following mixture can be separate using decantation method?
1 answer:
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Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
Answer:
The force is
Explanation:
The diagram for this question is shown on the first uploaded image
At Equilibrium the summation of the of force on the vertical axis is zero
i.e
=>
is the is the speed of water at the nozzle which can be mathematically evaluated as
substituting for R and
for
is the is the speed of water at the pipe which can be mathematically evaluated as
substituting for R and
for
is he density of water with value
Substituting values into the equation above
At Equilibrium the summation of the of force on the horizontal axis is zero
i.e
=>
Since The speed at both A and B nozzle are the same then remains the same
Substituting values
=>
Hence the force acting on the flange bolts required to hold the nozzle in place is
Answer:
Total contraction on the Bar = 1.22786 mm
Explanation:
Given that:
Total Length for aluminum bar = 600 mm
Diameter for aluminum bar = 40 mm
Hole diameter = 30 mm
Hole length = 100 mm
elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²
compressive load P = 180 KN = 180 × 10³ N
Calculate the total contraction on the bar = ???
The relation used in calculating the contraction on the bar is:
The relation used in calculating the total contraction on the bar can be expressed as :
Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)
i.e
Total contraction on the Bar =
Let's find the area of cross section without the hole and with the hole
Area of cross section without the hole is :
Using A = πd²/4
A = π (40)²/4
A = 1256.64 mm²
Area of cross section with the hole is :
A = π (40²-30²)/4
A = 549.78 mm²
Total contraction on the Bar =
Total contraction on the Bar =
Total contraction on the Bar = 2.117( 0.398 + 0.182)
Total contraction on the Bar = 2.117*(0.58)
Total contraction on the Bar = 1.22786 mm