Answer:
hello the diagram attached to your question is missing attached below is the missing diagram
answer :
a) 48.11 MPa
b) - 55.55 MPa
Explanation:
First we consider the equilibrium moments about point A
∑ Ma = 0
( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0
therefore ;<em> Fbd = 36 ( cos ∅tan30° - sin∅ ) kN ----- ( 1 )</em>
A ) when ∅ = 0
Fbd = 20.7846 kN
link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation
A = ( b - d ) t
b = 12 mm
d = 36 mm
t = 18
therefore loading area ( A ) = 432 mm^2
determine the maximum value of average normal stress in link BD using the relation below
бbd = 
= 20.7846 kN / 432 mm^2 = 48.11 MPa
b) when ∅ = 90°
Fbd = -36 kN
the negativity indicate that the loading direction is in contrast to the assumed direction of loading
There is compression in link BD
next we have to calculate the loading area using this equation ;
A = b * t
b = 36mm
t = 18mm
hence loading area = 36 * 18 = 648 mm^2
determine the maximum value of average normal stress in link BD using the relation below
бbd = = -36 kN / 648mm^2 = -55.55 MPa
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer : 9.36ohms/ temperature
Explanation:
Expression for the variation of resistance of platinum with temperature
Rt= Ro(1+*t)
Rt= resistance @ t°C
Ro= resistance @ 0°C
*= temperature coefficient of resistance
Calculate the change in resistance by putting 120ohms for Ro,
0.0039/K for *
20°C for t
Using this formula:
Rt = Ro(1+*t)
Rt- Ro = Ro*t
= (120ohms)(0.0039/K)(20°C)
= 9.36ohms/K
Answer:
D) AND gate.
Explanation:
Given that:
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print
These conditions are:
1. The printer's electronic circuits must be energized.
2. Paper must be loaded and ready to advance.
3. The printer must be "on line" with the microprocessor.
Now; if these conditions are met the logic gate produces a HIGH output indicating readiness to print.
The objective here is to determine the basic logic gate used in this circuit.
Now;
For NOR gate;
NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.
For NOT gate.
NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.
Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".
Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.
Answer:
0.867
Explanation:
The driver population factor (
)can be estimated using the equation below:
The value of the heavy vehicle factor (
) is determined below:
The values of the 
= 2 and 
= 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:
= 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833
Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:
= 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867
