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kotegsom [21]
2 years ago
3

If you need to drive during rush hour, try _____. A. changing lanes more often B. passing any vehicle that passes you C. staying

behind police vehicles, which move faster D. taking an alternate route with less traffic
Engineering
2 answers:
algol [13]2 years ago
6 0

The correct answer is D. Taking an alternate route with less traffic.

Explanation

The rush hour is a term used to refer to the period in which traffic is busiest by people, usually, the rush hour is in the morning when people take the transportation to go to their jobs and places of study and in the afternoons when they return home. During this time, most people are in massive means of transport such as the subway and buses, however, other people use their own means of transport such as motorcycles, bicycles, and cars. In the case of using a car, it is advisable to use alternate routes to avoid being in the places with the highest concentration of cars to avoid possible accidents with other cars and arrive earlier at the destination. So, the correct answer is D. Taking an alternate route with less traffic.

Anestetic [448]2 years ago
5 0
I believe the correct answer is D.
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49. A solenoid coil with a resistance of 30 ohms and an inductance of 200 milli-henrys, is connected to a 230VAC, 50Hz supply. D
Leya [2.2K]

Answer:

69.59 ohms

Explanation:

Given that

L=200\ mH

F=50\ hZ

R=30\ ohms

For inductance

X_L=2\times \pi\times f \times LX_L\\=2\times \pi \times 50\times 200\times 10^{-3}\\=62.8\ ohm

R=30\ ohmsImpedance\\Z=\sqrt{R^2+X_L^2}\\Z=\sqrt{30^2+62.8^2}\ ohms\\Z =69.59\ ohms

8 0
2 years ago
Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire
Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0
2 years ago
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
Outline an algorithm in **pseudo code** for checking whether an array H[1..n] is a heap and determine its time efficiency.
svlad2 [7]

Answer:

Condition to break: H[j] \geq max {H[2j] , H[2j+1]}

Efficiency: O(n).

Explanation:

Previous concepts

Heap algorithm is used to create all the possible permutations with K possible objects. Was created by B. R Heap in 1963.

Parental dominance condition represent a condition that is satisfied when the parent element is greater than his children.

Solution to the problem

We assume that we have an array H of size n for the algorithm.

It's important on this case analyze the parental dominance condition in order to the algorithm can work and construc a heap.

For this case we can set a counter j =1,2,... [n/2] (We just check until n/2 since in order to create a heap we need to satisfy minimum n/2 possible comparisionsand we need to check this:Break condition: [tex]H[j] \geq max {H[2j] , H[2j+1]}

And we just need to check on the array the last condition and if is not satisfied for any value of the counter j we need to stop the algorithm and the array would not a heap. Otherwise if we satisfy the condition for each j =1,2,.....,[n/2]p then we will have a heap.

On this case this algorithm needs to compare 2*(n/2) times the values and the efficiency is given by O(n).

3 0
2 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
2 years ago
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