All categories

3 years ago

A(n) ______ is used to measure fluid flow in engineering

Engineering

1 answer:

Kazeer [188]3 years ago

6 0

Explanation:

instrument of engineering and management skills that can be used to identify a device on a network of devices

You might be interested in

Define the Problem

gtnhenbr [62]

Answer:

rbrnrifnfnrfbdjrbfbfjrn

4 0

3 years ago

one number is 11 more than another number. find the two number if three times the larger number exceeds four times the smaller n

vaieri [72.5K]

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

$\begin{bmatrix}1 & -1 & 11\\3 & -4 & 4 \end{bmatrix}$

$\begin{bmatrix}1 & 0 & 40\\0 & 1 & 29 \end{bmatrix}$

a = 40

b = 29

6 0

3 years ago

Hỗ trợ mình với được không các bạn

Leya [2.2K]

Answer:

Explanation:

Be bop

6 0

3 years ago

An approach to a signalized intersection has a saturation flow rate of 1800 veh/h. At the beginning of an effective red, there a

Tcecarenko [31]

Answer: The total vehicle delay is

39sec/veh

Explanation: we shall define only the values that are important to this question, so that the solution will be very clear for your understanding.

Effective red time (r) = 25sec

Arrival rate (A) = 900veh/h = 0.25veh/sec

Departure rate (D) = 1800veh/h = 0.5veh/sec

STEP1: FIND THE TRAFFIC INTENSITY (p)

p = A ÷ D

p = 0.25 ÷ 0.5 = 0.5

STEP 2: FIND THE TOTAL VEHICLE DELAY AFTER ONE CYCLE

The total vehicle delay is how long it will take a vehicle to wait on the queue, before passing.

Dt = (A × r^2) ÷ 2(1 - p)

Dt = (0.25 × 25^2) ÷ 2(1 - 0.5)

Dt = 156.25 ÷ 4 = 39.0625

Therefore the total vehicle delay after one cycle is;

Dt = 39

4 0

3 years ago

Consider the circuit below where R1 = R4 = 5 Ohms, R2 = R3 = 10 Ohms, Vs1 = 9V, and Vs2 = 6V. Use superposition to solve for the

VladimirAG [237]

Answer:

The value of v2 in each case is:

A) V2=3v for only Vs1

B) V2=2v for only Vs2

C) V2=5v for both Vs1 and Vs2

Explanation:

In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.

Also, what the problem asks is the value V2 in each case, where:

$V_2=I_2R_2=V_{ab}$

If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.

In the first case we can use an equivalent resistance between R2 and R3:

$V_{ab}'=I_1'R_{2||3}=I_1'\cdot(\frac{1}{R_2}+\frac{1}{R_3})^{-1}$

And

$V_{S1}-I_1'R_1-I_1'R_4-I_1'R_{2||3}=0 \rightarrow I_1'=0.6A$

$V_{ab}'=I_1'R_{2||3}=3V=V_{2}'$

In the second case we can use an equivalent resistance between R2 and (R1+R4):

$V_{ab}''=I_3'R_{2||1-4}=I_3'\cdot(\frac{1}{R_2}+\frac{1}{R_1+R_4})^{-1}$

And

$V_{S2}-I_3'R_3-I_3'R_{2||1-4}=0 \rightarrow I_3'=0.4A$

$V_{ab}''=I_3'R_{2||1-4}=2V$

If we consider both batteries:

$V_2=I_2R_2=V_{ab}=V_{ab}'+V_{ab}''=5V$

7 0

3 years ago