If i would visit one of my friends it’s c
Answer:
Technician B
Explanation:
Technician B is correct in his argument. This is because according to what he said, as the computer pulses stimuli the coil will turn on and off, promoting an increase in the voltage that will cause the fluctuation. Technician A is incorrect because the procedure he indicated imposes that the voltage is checked at the negative terminal and not at the positive.
Answer:
True
Explanation:
Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true
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:
Using the formula
V =20y/(x^2+y^2)^1/2 - 20x/(x^2+y^2)^1/2
Hence fluid speed at x axis =20x/(x^2+y^2)^1/2
While the fluid speed at y axis =20y/(x^2+y^2)^1/2
Now the angle at 1, 5
We substitute into the formula above
V= 20×5/(1+25)^1/2= 19.61
For x we have
V = 20× 1/(1+25)^1/2= 3.92
Angle = 19.61/3.92= 5.0degrees
Angel at 5, and 2
We substitute still
V = 20×5/(2+25)^1/2=19.24
At 2 we get
V= 20×2/(2+25)^1/2=7.69
Dividing we get 19.24/7.69=2.5degrees
At 1 and 0
V = 20/(1)^1/2=20
At 0, v =0
Angel at 2 and 0 = 20degrees
At 5 and 2
V = 100/(25+ 4)^1/2=18.56
At x = 2
40/(√29)=7.43
Angle =18.56/7.43 = 2.49degrees.