A harmonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon.

a. What is the energyon in terms of k adm
b. If the oscillator is a bonded atom with k = 15 N/m and m = 4 × 10-26 kg, what is the frequency (Hz) of the emitted photon? (Note: the energy of a photon is Ephoton= hf)
c. In which region of the electromagnetic spectrum (x-ray, visible, microwave, etc.) does this photon belong?

Answer:

[tex] \mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}[/tex]

Explanation:

For this case we need to remember first thet the torque T is defined as:

[tex] T = FR[/tex]

Where T represent the torque, F the force acting in the inner cylinder and R the radius for the inner cylinder.

For the inner cylinder the force acting can be expressed as:

[tex] F = \mu A \frac{v}{l}[/tex]

Where [tex] \mu[/tex] represent the viscosity of the fluid, A the area of the inner cylinder, v represent the tangential velocity and l the thickness of fluid between the two cylinders.

And the tangential velocity for this case can be esxpressed as [tex] v = wR[/tex]

The info given is:

[tex] l = 0.035 in *\frac{1ft}{12in}=0.00292 ft[/tex]

[tex] R= \frac{D}{2} =\frac{6 in}{2}= 3 in*\frac{1ft}{12 in}=0.25 ft[/tex]

[tex] L = 5 ft[/tex] from the info given

N= 250 rpm represent the reveolutions per minute

[tex] T = 1.2 lbf ft[/tex] represent the torque given

We can find the surface area for the cylinder with this formula:

[tex] A= 2\pi R L[/tex]

And if we replace we got:

[tex] A= 2\pi 0.25 ft *5 ft= 7.85 ft^2[/tex]

Now we can find the tangential velocity like this:

[tex] v=wR= \frac2\pi *250 rpm* \frac{1min}{60s} * 0.25 ft=6.55\frac{ft}{s}[/tex]

Now we can set up the following equation for the torque:

[tex] T = FR[/tex]

[tex] T = \mu A \frac{v}{l} R[/tex]

And we can find the value for the viscosity [tex]\mu[/tex] like this:

[tex] \mu = \frac{T}{A \frac{v}{l} R}[/tex]

And if we replace we got:

[tex] \mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}[/tex]

Answer:

(a) Aluminum Indium Gallium Phosphide (AlInGaP). Band gap = 1.81eV  ≈ 2eV

(b) Gallium Nitride (GaN). Band Gap = 3.4eV

(c) Aluminium Gallium Arsenide (AlGaA). Band Gap = 1.42eV ≈ 2.16eV

(d) Zinc Selenide (ZnSe). Band Gap = 2.82eV

(e) Gallium Phosphide (GaP). Band Gap = 2.24eV

Explanation:

LED's are semi-conducting materials that convert electrical energy to light energy. The light color emitted from the LED depends on the semi-conducting material and other compositions.

The band gap of the semi-conductor determines its wavelength. High band gap semi-conductors emit lower wavelengths which means greater power(UV semi-conducting macterials fall under this category).

To calculate the theoretical density of Niobium with a BCC structure, we use its atomic radius and atomic weight, converting these into the cube's edge length using the BCC relation, then apply the density formula.

To calculate the theoretical density of Niobium (Nb), which has a body-centered cubic (BCC) crystal structure, we first use the known values: atomic radius = 0.143 nm (or 0.143 × 10-9 m) and atomic weight = 92.91 g/mol. The formula for the density (ρ) in a BCC structure is ρ = (2 × M) / (a3 × NA), where M is the atomic mass, a is the edge length of the cube, and NA is Avogadro's number (6.022 × 1023 atoms/mol).

Since it's a BCC structure, the atomic radius relates to the cube's edge length (a) as a = 4r / √3. Substituting the given atomic radius, we find a = 4 * 0.143 × 10-9 m / √3. Then, to find the density, we substitute M (92.91 g/mol), a, and NA into the density formula. This calculation will give us the theoretical density of Niobium in g/cm3.

a. It's not entirely reasonable to generalize the conclusion to all 18-year-olds with publicly accessible MySpace profiles due to the study's limited sample size and potential biases.

b. b. No, it's not reasonable to generalize the conclusion to all 18-year-old MySpace users because not all users have publicly accessible profiles.

c. While somewhat reasonable, generalizing to MySpace users with publicly accessible profiles requires caution, considering potential individual differences within this subgroup.

The study only looked at a random sample of 500 profiles, which might not be fully representative of all 18-year-olds on MySpace. Generalizing to all 18-year-old MySpace users is not reasonable as not all users have publicly accessible profiles, which could skew the findings.  

Additionally, factors such as cultural differences, regional variations, and individual preferences could influence whether adolescents choose to display sport or hobby involvement on their profiles.

See text

The article "Display of Health Risk Behaviors on MySpace by Adolescents" (Archives of Pediatrics and Adolescent Medicine [2009]:

) described a study in which researchers looked at a random sample of 500 publicly accessible MySpace web profiles posted by 18-year-olds. The content of each profile was analyzed. One of the conclusions reported was that displaying sport or hobby involvement was associated with decreased references to risky behavior (sexual references or references to substance abuse or violence).

a. Is it reasonable to generalize the stated conclusion to all 18-year-olds with a publicly accessible MySpace web profile? What aspect of the study supports your answer?

b. Not all MySpace users have a publicly accessible profile. Is it reasonable to generalize the stated conclusion to all 18-year-old MySpace users? Explain.

c. Is it reasonable to generalize the stated conclusion to all MySpace users with a publicly accessible profile? Explain.

The calculation of the exit temperature of water in the heated pipe involves using the energy balance equation, considering the heat lost through the pipe walls by convection, and then finding the change in thermal energy of the water via the heat transfer equation to solve for the exit temperature.

Exit Temperature of Water in a Heated Pipe

To determine the exit temperature of water at the end of an over-pressurized heated pipe, we must consider the energy balance for the water flowing through the pipe. Based on the first law of thermodynamics, the change in thermal energy of the water will be equal to the heat lost through the pipe walls by convection:

Q = mc_p
(Exit Temperature - Inlet Temperature)

In this case, Q will be negative, since the water is losing heat to the surrounding air. The heat transfer from the pipe to the air is given by:

Q = hA(T_surface - T_air)

The area A for heat transfer is the external surface area of the pipe (
2
classes Math.PI
* radius * length of the pipe). Since we have the heat transfer coefficient h, the surface temperature of the pipe T_surface, and the air temperature T_air, we can calculate Q. Then we can use the mass flow rate m and the constant pressure specific heat c_p to find the exit temperature of the water.

To solve, we first calculate Q, then rearrange the first equation to solve for the Exit Temperature.

Answer:

(a) The charges of 75 C must flow through battery.

(b) The electrons must flow from a to b.

Explanation:

(a)

The relationship between the energy and voltage is as follows:

Energy = (Voltage)(Current)(Time)

but, (current)(time) = charge

therefore,

Energy = (Voltage)(Charge)

Charge = (Energy)/(Voltage)

Charge = (900 J)/(12 V)

Charge = 75 C

(b)

Since, Vab = Va - Vb = 12 V

Hence, the equation suggest that Va is at higher potential then Vb.

As, the current flows from higher to lower potential.

Thus, electrons must flow from a to b

Answer:

a) 0.759 m , 0.759 m

b) 10.1 m , 10.3 m

c) 12.3 m , 13 m

Explanation:

Vapour Pressure included:

[tex]P_{atm,vp} = y*h + P_{vp} \\\\h = \frac{P_{atm,vp}-P_{vp}}{y} ... Eq1[/tex]

mercury

[tex]h_{Hg,vp} = \frac{101*10^3-1.6*10^(-1)}{133*10^3} \\\\h_{Hg,vp} = 0.759m[/tex]

[tex]h_{Hg} = \frac{101*10^3}{133*10^3} \\\\h_{Hg} = 0.759m[/tex]

water

[tex]h_{H2O,vp} = \frac{101*10^3-1.77*10^3}{9.8*10^3} \\\\h_{H2O,vp} = 10.1m[/tex]

[tex]h_{H2O} = \frac{101*10^3}{9.8*10^3} \\\\h_{H20} = 10.3m[/tex]

Ethyl Alcohol

[tex]h_{EA,vp} = \frac{101*10^3-5.9*10^3}{7.74*10^3} \\\\h_{EA,vp} = 12.3m[/tex]

[tex]h_{EA} = \frac{101*10^3-5}{7.74*10^3} \\\\h_{EA} = 13m[/tex]

For mercury barometers the effects of vapour pressure are negligible and required height for mercury barometer is reasonable as compared to that of water and ethyl alcohol.

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

 Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

 Eq 3

Finally, solving for unknown pressure as a function of hoop strain:

 Eq 4

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

 Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

 Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

 Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

 Eq 8

For Wheat stone bridge:

 Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

Answer:

in situ treatment would be better if; (1) a large volume of soil is to be treated at once, and (2) a permeable sandy soil (un-compacted) is to be treated.

ex situ treatment would be better would be better if; (1) a wider range of contaminants and soil types are to be treated, and (2) there is more certainty about the uniformity of treatment because of the ability to homogenize, screen, and continuously mix the soil.

Explanation:

in situ treatment would be better if;

ex situ treatment would be better would be better if;

Answer:

bypassed fraction B will be B= 0.105 (10.5%)

Explanation:

doing a mass balance of SO₂ at the exit

total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂

F*(1-er) =  Fs*(1-es) + Fb

where

er= required efficiency

es= scrubber efficiency

Fs and Fb = total mass inflow of  SO₂ to the scrubber and to the bypass respectively

F= total mass inflow of  SO₂

and from a mass balance at the inlet

F= Fs+ Fb

therefore the bypassed fraction B=Fb/F is

F*(1-er) =  Fs*(1-es) + Fb

1-er= (1-B)*(1-es) +B

1-er = 1-es - (1-es)*B + B

(es-er) = es*B

B= (es-er)/es = 1- er/es

replacing values

B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)

Answer:

The Java code is given below with appropriate variable names and tags for better understanding

Explanation:

import java.util.Scanner;

public class LabProgram{

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int highwayNumber;

       int primaryNumber;

       highwayNumber = scnr.nextInt();

       if(highwayNumber<1 || highwayNumber>999)

           System.out.println(highwayNumber+" is not a valid interstate highway number.");

       else{

           if(highwayNumber>=1 && highwayNumber<=99){

               System.out.print("The "+highwayNumber+" is primary, going ");

               if(highwayNumber%2==1)

                   System.out.println("north/south.");

               else

                   System.out.println("east/west.");

           }

           else{

               primaryNumber = highwayNumber%100;

               System.out.print("The "+highwayNumber+" is auxillary, serving the "+primaryNumber+", going ");

               if(primaryNumber%2==1)

                   System.out.println("north/south.");

               else

                   System.out.println("east/west.");

           }

       }

   }

}

The program is an illustration of conditional statements.

Conditional statements are statements whose execution is dependent on its truth value.

The missing code segment in Java where comments are used to explain each line is as follows:

import java.util.Scanner;

public class Main {

   public static void main(String [] args){  

       Scanner scnr = new Scanner(System.in);  

       int highwayNumber; int primaryNumber;

       highwayNumber = scnr.nextInt();

       //This checks if the highwayNumber is not between 1 and 999 (inclusive)

       if (highwayNumber <1 || highwayNumber > 999){

           //If yes, the highwayNumber is invalid

           System.out.print(highwayNumber+" is not a valid interstate highwayNumber number.");

       }

       //If otherwise

       else{

           //This checks if highwayNumber is less than 100

           if (highwayNumber< 100){

               if (highwayNumber%2 == 0){

                   //Even highwayNumber are primary going east/west

                   System.out.print("I-"+highwayNumber+" is primary, going east/west.");

               }

           else{

               //Odd highwayNumber are primary going north/south

               System.out.print("I-"+highwayNumber+" is primary, going north/south.");

           }

           }

           //Otherwise

           else{

               if ((highwayNumber%100) % 2 == 0){

                   //Even highwayNumber are auxiliary going east/west

                   System.out.print("I-"+highwayNumber+" is auxiliary, going east/west.");

               }

               else{

                   //Even highwayNumber are auxiliary going north/south

                   System.out.print("I-"+highwayNumber+" is auxiliary, going north/south.");

               }

           }

       }

   }

}

Read more about similar programs at:

https://brainly.com/question/22078310

Answer:

(a) Mass flow rate is 2528.75 kg/min

(b) Molar flow rate is 458.1 mol/s

(c) Toulene is in-compressible at standard temperature and pressure

Explanation:

The volume flow rate of liquid toulene is given:

Volume Flow Rate = 175 m^3/h

Density of Liquid Toulene = 867 kg/m^3

Molar mass of Toulene = 92 g/mol = 0.092 kg/mol)

(a)

Thus, the mass flow rate of toulene will be:

Mass Flow Rate = (Volume Flow Rate)(Density)

Mass Flow Rate = (175 m^3/h)(867 kg/m^3)

Mass Flow Rate = (151725 kg/h)(1 h/60 min)

Mass Flow Rate = 2528.75 kg/min

(b)

Now, for molar flow rate we use formula:

Molar Flow Rate = (Mass Flow Rate)/(Molar Mass)

Molar Flow Rate = (2528.75 kg/min)/(0.092 kg/mol)

Molar Flow Rate = (27486.41 mol/min)(1 min/60sec)

Molar Flow Rate = 458.1 mol/s

(b)

The assumption is that Toulene is in-compressible at standard temperature and pressure. So, that its density can be taken constant.

Answer:

v = 1779 veh/h

Explanation:

We will calculate the peak hour volume by using the following formula:-

Vp = v / ( PHF)(N)(Fg)(Fhv)

here,

1. v is the peak hour volume

2. vp is the analysis flow rate

3. PHF is the peak hour factor

4. N is the number of lanes

5. Fg is the grade adjustment factor which is 0.99 for rolling terrain > 1200 pc/h/ln

6. Fhv is the heavy vehicle adjustment factor

vp = v / (PHF) (N) (Fg) (Fhv)

1250 = v / (0.84)(2)(1)(0.847)

v = 1779 veh/h

The time that the maximum queue length occurs would be c. 49.4 min

To find the time at which the maximum queue length occurs, we need to determine when the arrival rate equals the departure rate.

The queue length increases when the arrival rate exceeds the departure rate and decreases when the departure rate exceeds the arrival rate. The maximum queue length occurs when the arrival rate equals the departure rate.

Substituting the given functions into the equation, we get:

[tex]\[ 1.8 + 0.25t - 0.0030t^2 = 1.4 + 0.11t \][/tex]

Rearranging the terms, we get a quadratic equation:

[tex]\[ -0.0030t^2 + 0.25t - 0.11t + 1.8 - 1.4 = 0 \][/tex]

[tex]\[ -0.0030t^2 + 0.14t + 0.4 = 0 \][/tex]

Now, we can solve this quadratic equation to find the value(s) of t at which the maximum queue length occurs. We can use the quadratic formula:

[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Where:

a = -0.0030

b = 0.14

c = 0.4

Calculate the values of t using this formula. Then, we'll choose the appropriate value based on the physical meaning of the problem.

Using the quadratic formula:

[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

[tex]\[ t = \frac{-0.14 \pm \sqrt{(0.14)^2 - 4(-0.0030)(0.4)}}{2(-0.0030)} \][/tex]

[tex]\[ t = \frac{-0.14 \pm \sqrt{0.0196 + 0.0048}}{-0.0060} \][/tex]

[tex]\[ t = \frac{-0.14 \pm 0.156}{-0.0060} \][/tex]

Now, we have two possible values for t :

1. [tex]\( t_1 = \frac{-0.14 + 0.156}{-0.0060} \)[/tex]

2. [tex]\( t_2 = \frac{-0.14 - 0.156}{-0.0060} \)[/tex]

Calculate these values:

1. [tex]\( t_1 = \frac{0.016}{-0.0060} = -2.67 \)[/tex]

2. [tex]\( t_2 = \frac{-0.296}{-0.0060} = 49.4[/tex]

Since t represents time, it cannot be negative. The maximum queue length occurs at approximately t = 49.4 minutes after the beginning of the observation.

Answer:

V = 56.8 mV

Explanation:

When a current I flows across a circuit element, if we assume that the dimensions of the circuit are much less than the wavelength of the power source creating this current, there exists a fixed relationship between the power dissipated in the circuit element, the current I and the voltage V across it, as follows:

P = V*I

By definition, power is the rate of change of energy, and current, the rate of change of the charge Q, so we can replace P and I, as follows:

E/t = V*q/t ⇒ E = V*Q

Solving for V:

V = E/Q = 94.2 mJ /1.66 C = 56.8 mV

Answer:

Program comments

Answer:

a) 69,630KW

b) 203 KW

Explanation:

The data obtained from Tables A-4, A-5 and A-6 is as follows:

[tex]h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2} = 255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\[/tex]

[tex]x_{4} = \frac{s_{4} - s_{f} }{s_{fg} } \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\[/tex]

The power produced and consumed by turbine and pump respectively are:

[tex]W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW[/tex]

Answer:

Fde = 1080 N (3 sig fig)

Explanation:

Taking point E

Sum of forces at E in x direction:

Fde cos (30) - Fcd = 0  ..... Eq 1

Sum of forces at E in y direction:

Fde sin (30) - Wf = 0   ..... Eq 2

Wf = m*g = 55 * 9.81 = 539.55 N

Using Eq 2 to evaluate Fde

Fde = Wf / sin (30) = 539.55 / sin (30) = 1079.1 N

Answer: Fde = 1080 N (3 sig fig)

Answer:

Time taken = 5.41mins

Explanation:

A step by step calculations with detailed explanation has been attached below.

Answer: attached below. rate brainliest if correct please

Answer with explanation:

As is the question the answer would be 19 hours, and the key to solving it is in the phrase in 15 more hours, basically what they are saying is that the small pipe takes 15 hours more than both the big and the small to fill the tank. Since both pipes working together can fill the tank in 4 hours we need to add 4 and 15 to solve the problem.

If the question is how many hours would it take to fill the tank using only the big pipe? Then we could solve t for the following equation:

[tex]\frac{1}{4+15} + \frac{1}{t} = \frac{1}{4}[/tex]

Getting as a result: 5.06

Note that the equation is the result of taking the rate of the small pipe (what we solved before), plus the unknown rate of the big one equals the rate of both.

Answer: c. Fully actuated

Explanation: fully actuated signals are completely influenced by volumes if traffic and employs sensors at all approaches for its detection.

Answer:

I am writing a function in C++                              

Explanation:

#include <iostream>

using namespace std;

int pyramid(int height) // function pyramid with parameter height

{int distance; //variable for spaces

   for(int i = 1, j = 0; i <= height; ++i, j= 0)  //handle the rows

   {

for(distance= 1; distance<= height-i; ++distance)

// handles the spaces & columns

    { cout <<"  ";        } //prints spaces between stars

       while(j!= 2*i-1)  //handles shape and spaces

       {   cout << "* "; // printing stars

           ++j;        }

       cout << '\n';    }   } // for the next line

int main()

{

int height; //declare height variable

cout <<"Enter height of pyramid "; //asks user to enter height of pyramid

cin>>height; //stores value of height

pyramid(height); //calls pyramid function

}

Answer:

Super insulation are obtained by using layers of highly reflective sheets separated by glass fibers in an vacuumed space. Radiation heat transfer between any of the surfaces is inversely proportional to the number of sheets used and thus heat lost by radiation will be very low by using these highly reflective sheets which will an effective way of heat transfer.

Explanation:

Answer:

(userVal < 0) ? "negative" : "non-negative" ;

Explanation:

The above code has been written using Java's ternary operator (? : ).

Java uses this operator to write conditional expressions that are similar to the regular if ... else statements.

This expression has three parts,

i. the conditional statement: this is the expression before the ? mark. In this case, (userVal < 0). This expression contains the condition to be tested for and it returns true or false.

ii. the second part is the statement after the ? mark. In this case, "negative". This expression will be executed if the conditional statement returns true.

iii. the third part is the statement after the : mark. In this case, "non-negative". This expression will be executed if the conditional statement returns false.

So in this case, if userVal < 0, the string "negative" will be evaluated. Otherwise, "non-negative".

Answer:

The score for both exams 88 is above the 90 percentile, so then Brynne qualified for honors. See the explanation below.

Explanation:

Assuming the following question:"At Westtown High School, the mean score on the French final examination was 81 with a standard deviation of 5, while the mean score on the Spanish final examination was 72 with a standard deviation of 12.

To earn a language honor at graduation, students must score in the 90th percentile on all their language final exams. Brynne scored 88 on both the French exam and the Spanish exam. Is Brynne qualified for honors?"

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

French case

Let X the random variable that represent the scores for the French exam of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(81,5)[/tex]  

Where [tex]\mu=81[/tex] and [tex]\sigma=5[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.1[/tex]   (a)

[tex]P(X<a)=0.90[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=1.28<\frac{a-81}{5}[/tex]

And if we solve for a we got

[tex]a=81 +1.28*5=87.4[/tex]

So the value for the scores that separates the bottom 90% of data from the top 10% is 87.4 (90th percentile).

And since the score of Brynne is 88 is above the 90 percentile  

Spanish case

Let X the random variable that represent the scores for the Spanish exam of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(72,12)[/tex]  

Where [tex]\mu=72[/tex] and [tex]\sigma=12[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.1[/tex]   (a)

[tex]P(X<a)=0.90[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=1.28<\frac{a-72}{12}[/tex]

And if we solve for a we got

[tex]a=72 +1.28*12=87.36[/tex]

So the value for the scores that separates the bottom 90% of data from the top 10% is 87.36 (90th percentile).

And since the score of Brynne is 88 is above the 90 percentile  

Answer: 62 cos(50t - 70°)

Explanation:

First we need to convert all sines into cosines because phasor forms are represented through cosine. For that we will use the fact that sin(wt + ∅) = cos(wt + ∅ - 90°)

Therefore, 37sin50t = 37cos(50t - 90°)

Now we have 37cos(50t - 90°)+ 30 cos(50t – 45°). We need to convert them into phasor form to add the terms. For that we will use the fact Acos(wt+∅)=A∠∅ which can be represented using real and imaginary parts as A [cos(∅)+jsin(∅)].

So,

37cos(50t - 90°)

= 37∠-90°

= 37[cos(-90°)+jsin(-90°)

=37[0+j(-1)]

= -j37

Similarly,

30 cos(50t – 45°)

=30∠-45

=30[cos(-45)+jsin(-45)

=30[0.707-j0.707]

=21.21 - j21.21

37cos(50t - 90)+ 30 cos(50t – 45°) = -j37+21.21 - j21.21 = 21.21 - j58.21

Now we need to convert the real and imaginary parts back to cosine form. We will first calculate the magnitude by the formula √a²+b² where a and b are the real and imaginary parts respectively.

Here a=21.21 and b=58.21

magnitude = √(21.21)²+(58.21)²=61.95≅62

For calculating phase ∅ the formula is ∅=inversetan (b/a) where a and b are the real and imaginary parts respectively.

∅=inversetan(-58.21/21.21)

= -69.9°≅-70°

So the final answer is 62cos(50t-70°)

The value of [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] is [tex]\(61.97 \cos(50t - 70^\circ)\)[/tex].

To solve [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] using phasors, we follow these steps:

Step-by-Step Calculation:

1. Express each term as a phasor:

- The term [tex]\(37 \sin 50t\)[/tex] :

    - Convert to cosine form: [tex]\(37 \sin 50t = 37 \cos(50t - 90^\circ)\)[/tex]

    - Phasor form: [tex]\(37 \angle -90^\circ\)[/tex]

  - The term [tex]\(30 \cos(50t - 45^\circ)\)[/tex] :

    - Phasor form: [tex]\(30 \angle -45^\circ\)[/tex]

2. Convert the phasors to rectangular form:

  - [tex]\(37 \angle -90^\circ\)[/tex] :

    [tex]\[ 37 \cos(-90^\circ) + j 37 \sin(-90^\circ) = 0 - j 37 = -j 37 \][/tex]

- [tex]\(30 \angle -45^\circ\)[/tex]:

    [tex]\[ 30 \cos(-45^\circ) + j 30 \sin(-45^\circ) = 30 \left(\frac{\sqrt{2}}{2}\right) - j 30 \left(\frac{\sqrt{2}}{2}\right) = 21.2132 - j 21.2132 \][/tex]

3. Add the rectangular components:

 [tex]\[ -j 37 + (21.2132 - j 21.2132) \][/tex][tex]\[ = 21.2132 - j (37 + 21.2132) \][/tex]

  [tex]\[ = 21.2132 - j 58.2132 \][/tex]

4. Convert the result back to polar form:

  - Magnitude:

   [tex]\[ R = \sqrt{21.2132^2 + 58.2132^2} = \sqrt{449.54 + 3388.78} = \sqrt{3838.32} = 61.97 \][/tex]

  - Angle:

   [tex]\[ \theta = \tan^{-1}\left(\frac{-58.2132}{21.2132}\right) = \tan^{-1}(-2.743) \approx -70^\circ \][/tex]

5. Express the final result:

  Using the positive magnitude, the expression in cosine form is:

  [tex]\[ 37 \sin 50t + 30 \cos(50t - 45^\circ) = 61.97 \cos(50t - 70^\circ) \][/tex]

Answer:

The resistance (R) of the resistor is 2.4 ohm

The maximum voltage (V) is 4.33V

The maximum current (I) is 1.80A

Explanation:

Resistance (R) = resistivity×length/area

Resistivity = 0.003 ohm meter, length = 2mm = 2/1000 = 0.002m, width = 1.25mm = 1.25/1000 = 0.00125m, height = 0.5mm = 0.0005m, area = width × height = 0.00125m × 0.0005m = 6.25×10^-7m^2

R = 0.003×0.002/6.25×10^-7 = 3.2 ohm

Power (P) = V^2/R

V^2 = P × R = 7.81 × 3.2= 24.992

V = √24.992 = 4.99V

P = IV

I = P/V = 7.81/4.99 = 1.57A

Answer:

a. Self declared by the manufacturer

Explanation:

Recycled content refers to the portion of materials used in a product that have been diverted from the solid waste stream. If those materials are diverted during the manufacturing process, they are be referred to as pre-consumer recycled content (sometimes referred to as post-industrial). If they are diverted after consumer use, they are

post-consumer .

Answer:

sulfur dioxide

Explanation:

The scrubber is an apparatus installed in a coal-fired power plant to clean the passing gas through the smokestack.  Due to the norm enacted through the clean air Act, almost all the scrubber used in the U.S is used to remove sulfur concentration from coal. it can remove approx 90-95% SO_2 from the smokestack.

Answer:

sulfur dioxide.

Explanation: