Select the correct answer What is the simplest definition of a manufacturing process?
2 answers:
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Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
where;
R= radius
= coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
= 0.20
So;
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
Answer:
The exact time when the sample was taken is = 0.4167337 hr
Explanation:
The diagram of a sketch of the tank is shown on the first uploaded image
Let A denote the first inlet
Let B denote the second inlet
Let C denote the single outflow from the tank
From the question we are given that the diameter of A is = 1 cm = 0.01 m
Area of A is =
=
Velocity of liquid through A = 0.2 m/s
The rate at which the liquid would flow through the first inlet in terms of volume = =
i.e is
=
The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume
=
= 0.016324
From the question the diameter of B = 2 cm = 0.02 m
Area of B =
Velocity of liquid through B = 0.01 m/s
The rate at which the liquid would flow through the first inlet in terms of volume = =
i.e is
=
The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume
=
= 0.00330642
From the question The flow rate in term of volume of the outflow at the time of measurement is given as = 0.5 L/s
And also from the question the mass of potassium chloride at the time of measurement is given as 13 g/L
So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume
=
= Note (1 Kg = 1000 g)
= 0.0065 kg/s
Considering potassium chloride
Let denote the rate at which liquid flows in terms of mass as as i.e change in mass with respect to time hence
Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )
(0.016324 + 0.00330642) - 0.0065 =
=> 0.01313122 t = (m - )
From the question (m - ) is given as = 19.7 Kg
Hence the time when the sample was taken is given as
0.01313122 t = 19.7 Kg
=> t = 1500.2414 sec
t = .4167337 hours (1 hour = 3600 seconds)
