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statuscvo [17]
3 years ago
8

PELASEE HELPPP WITH MARK BRAINLIST!!!! You are stopped at a red light, and a long string of cars is crossing in front of you. Wh

en their traffic light turns red and yours turns green, the crossing traffic continues to enter the intersection. You should
Remain stopped until all the crossing traffic has crossed through the intersection.

Call the police on your cell phone to report them.

Move forward while honking your horn, to force the crossing traffic to obey the traffic signal.
Engineering
2 answers:
choli [55]3 years ago
7 0

Answer:

1st one.

Explanation:

I think that because they the one who is going to get a ticket and also you will not gonna get in a car accident.

Tpy6a [65]3 years ago
3 0

Answer:

Hello there the Correct answer is A.

Because it shows that you can GO, but you must first let any vehicles, bicycles, or pedestrians remaining in the intersection get through before you move ahead. You can turn left ONLY if you have enough space to complete the turn before any oncoming vehicle, bicycle, or pedestrian becomes a hazard.

You might be interested in
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
3 years ago
On-site oil storage containers must be marked "Used Oil."<br><br> True. <br><br> False.
Scrat [10]

Answer:

How must used oil storage containers be marked? Containers and aboveground tanks used to store used oil at generator facilities must be labeled or marked clearly with the words “Used Oil" (40 CFR Section 279.22(c)).

Explanation:

i think it will help you

7 0
3 years ago
Question 1: Final Results = What are the values of the resistances such that the gain = -100, Rin = 1 MI2. Don't use resistances
lidiya [134]

Answer:

Explanation:

In a study of algebra, you will encounter many families of equations, or groups of

equations that share common characteristics. Of interest to us here is the family of

linear equations in one variable, a study that lays the foundation for understanding

more advanced families. In addition to solving linear equations, we’ll use the skills we

develop to solve for a specified variable in a formula, a practice widely used in science,

business, industry, and research.

A. Solving Linear Equations Using Properties of Equality

An equation is a statement that two expressions are

equal. From the expressions and

we can form the equation

which is a linear equation in one variable. To solve

an equation, we attempt to find a specific input or xvalue that will make the equation true, meaning the

left-hand expression will be equal to the right. Using

Table 1.1, we find that is a

true equation when x is replaced by 2, and is a false

equation otherwise. Replacement values that make

the equation true are called solutions or roots of the equation.

4 0
2 years ago
This method will sell the seat in row i and column j unless it is already sold. A ticket is sold if the price of that seat in th
blsea [12.9K]

Answer:

The solution code is written in Java.

  1. public class Movie {
  2.    private double  [][] seats = new double[5][5];
  3.    private double totalSales;
  4.    public Movie(){
  5.        for(int i= 0; i < this.seats.length; i++){
  6.            for(int j = 0; j < this.seats[i].length; j++){
  7.                this.seats[i][j] = 12;
  8.            }
  9.        }
  10.        this.totalSales = 0;
  11.    }
  12.    public boolean bookSeat(int i, int j)
  13.    {
  14.        if(this.seats[i][j] != 0){
  15.            this.totalSales += this.seats[i][j];
  16.            this.seats[i][j] = 0;
  17.            return true;
  18.        }else{
  19.            return false;
  20.        }
  21.    }
  22. }

Explanation:

The method, bookSeat(), as required by the question is presented from Line 16 - 26 as part of the public method in a class <em>Movie</em>.  This method take row,<em> i</em>, and column,<em> j</em>, as input.

By presuming the seats is an two-dimensional array with all its elements are  initialized 12 (Line 7 - 10). This means we presume the movie ticket price for all the seats are $12, for simplicity.

When the<em> bookSeat() </em>method is invoked, it will check if the current price of seats at row-i and column-i is 0. If not, the current price, will be added to the <em>totalSales </em>(Line 19)<em> </em>and then set the price to 0 (Line 20) and return <em>true</em> since the ticket is successfully sold (Line 21).  If it has already been sold, return <em>false</em> (Line 23).

8 0
3 years ago
For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or
Alborosie

Answer:

Glass: Low-Loss dielectric

  α = 8.42*10^-11 Np/m

  β = 468.3 rad/m

  λ = 1.34 cm

  up = 1.34*10^8 m/s

  ηc = 168.5 Ω

Tissue: Quasi-Conductor

  α = 9.75 Np/m

  β = 12.16 rad/m

  λ = 51.69 cm

  up = 0.52*10^8 m/s

  ηc = 39.54 + j 31.72 Ω        

Wood: Good conductor

  α = 6.3*10^-4 Np/m

  β = 6.3*10^-4 Np/m

  λ = 10 km

  up = 0.1*10^8 m/s

  ηc = 6.28*( 1 + j )

Explanation:

Given:

Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz

Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.

Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz

Find:

Determine if  the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then  calculate α, β, λ, up, and ηc:

Solution:

- We need to determine the loss tangent to determine category of the medium as follows:

                                σ / w*εr*εo

Where, w is the angular speed of wave

            εo is the permittivity of free space = 10^-9 / 36*pi

- Now we classify as follows:

    Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\  

- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.

    Glass: Low-Loss dielectric

    Tissue: Quasi-Conductor

    Wood: Good conductor

- Now we will use categorized material base equations from Table 17-1 as follows:

     Glass: Low-Loss dielectric

          α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)

          α = 8.42*10^-11 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)

          β = 468.3 rad/m

          λ = 2*pi / β = 2*pi / 468.3

          λ = 1.34 cm

          up = λ*f = 0.0134*10^10

          up = 1.34*10^8 m/s

          ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)

          ηc = 168.5 Ω

     Tissue: Quasi-Conductor

          α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)

          α = 9.75 Np/m

          β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)

          β = 12.16 rad/m

          λ = 2*pi / β = 2*pi / 12.16

          λ = 51.69 cm

          up = λ*f = 0.5169*100*10^6

          up = 0.52*10^8 m/s

          ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5

          ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

          ηc = 39.54 + j 31.72 Ω

     Wood: Good conductor

          α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )

          β = α = 6.3*10^-4 Np/m

          λ = 2*pi / β = 2*pi / 6.3*10^-4

          λ = 10 km

          up = λ*f = 10,000*1*10^3

          up = 0.1*10^8 m/s

          ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4

          ηc = 6.28*( 1 + j )

         

           

         

8 0
3 years ago
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