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Luba_88 [7]
2 years ago
7

A flow rate sensing device used on a liquid transport pipeline functions as follows. The device provides a 5-bit output where al

l five bits are zero if the flow rate is less than 10 gallons per minute. The first bit is 1 if the flow rate is at least 10 gallons per minute; the first and second bits are 1 if the flow rate is at least 20 gallons per minute; the first, second, and third bits are 1 if the flow rate is at least 30 gallons per minute; and so on. The five bits, represented by the logical variables A, B, C, D, and E, are used as inputs to a device that provides two outputs Y and Z.
a. Write an equation for the output Y if we want Yto be 1 iff the flow rate is less than 30 gallons per minute.
b. Write an equation for the output Z if we want Z to be 1 iff the flow rate is at least 20 gallons per minute but less than 50 gallons per minute.

Engineering
1 answer:
marysya [2.9K]2 years ago
8 0

Answer:

Explanation:

The step by step analysis is as shown in the attached files.

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What is the damped natural frequency (in rad/s) of a second order system whose undamped natural frequency is 25 rad/s and has a
TiliK225 [7]

Answer:

damped natural frequency = 23.84 rad/s

Explanation:

given data

damping ratio = 0.3

undamped natural frequency = 25 rad/s

to find out

damped natural frequency of a second order system

solution

we know that if damping ratio is = 0

then it is undamped system

and if damping ratio is > 1

then it is overdamped system

and  and if damping ratio is ≈ 1

then it is critical damped system

so damped natural frequency of a second order system formula is

damped natural frequency = wn × \sqrt{1-r^2}

here wn is undamped natural frequency and r is damping ratio

damped natural frequency = 25 × \sqrt{1-0.3^2}

damped natural frequency = 23.84 rad/s

3 0
2 years ago
A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a
Fed [463]

Answer:

<em>minimum required diameter of the steel linkage is 3.57 mm</em>

<em></em>

Explanation:

original length of linkage l = 10 m

force to be transmitted  f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = 2*10^{8}  N/m^{2}

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/(2*10^{8}) = 10^{-5} m^2

recall that area = \pi d^{2} /4

10^{-5} = \frac{3.142*d^{2} }{4} = 0.7855d^{2}

d^{2} = \frac{10^{-5} }{0.7855} = 1.273*10^{-5}

d = \sqrt{1.273*10^{-5}  } = 3.57*10^{-3} m = 3.57 mm

<em>maximum diameter of  the steel linkage d = 3.57 mm</em>

4 0
2 years ago
Suppose we store a relation R (x,y) in a grid file. Both attributes have a range of values from 0 to 1000. The partitions of thi
leva [86]

Answer:

For (a) The total number of buckets from the given query for the relation is 25 buckets (b) the nearest neighboring query is (80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

Explanation:

From the question stated, we need to define what a Grid file is

Grid File it is a structure of data that are used to divide the total space into a grid non-periodic, where set of point (small) are defined by more than one cells of the grid.

(a)Finding buckets for the query

The relation is divided into two parts which ranges from 0 to 1000, the first part is partitioned in every 20 units, at 20, 40, 60 etc; a second part is partitioned into every 50 units at 50, 100, 150 etc.

The total number of buckets from the given query for the relation is 25 buckets

(b)Finding the closest point or nearest point

The closest point discovered in the distance is little above 15

These points are are the points closer to the point target (110, 205) which can be found in five neighboring rectangles with left corners lower is stated as follows:

(80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

3 0
2 years ago
An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi
Fittoniya [83]

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.

3 0
2 years ago
Your task is to fill in the missing parts of the C code to get a program equivalent to the generated assembly code. Recall that
Rudik [331]

Answer:

See Explaination

Explanation:

//Function

long loop (long x, long n)

{

//Declare a variable named result and initialize it to zero

long result = 0;

//Declare a variable named mask

long mask;

//For loop

for(mask = 1; mask != 0; mask = mask << (n & 0xFF))

{

//Calculate

result | = (x&mask);

}

//Return result

return result;

}

6 0
2 years ago
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