1. A group of 45 tests on a given type of concrete had a mean strength of 4780 psi and a standard deviation of 525 psi. Does this concrete satisfy the requirements of ACI code for 4000 psi concrete

Answers

Answer 1

Answer:

Yes, because both design compressive stress is greater than 4000 psi

Explanation:

To design a concrete that satisfy the requirements of ACI code for 4000 psi concrete

Step 1: design the compressive stress, using the two equations below

[tex]f_{c} =f_{cr}-1.34*s[/tex] -------equation 1

where;

[tex]f_{c}[/tex] is the design compressive stress

[tex]f_{cr}[/tex] is the critical stress = 4780 psi mean strength

s is the standard deviation = 525 psi

[tex]f_{c} = 4780 -1.34*525[/tex] = 4076.5 psi

Step 2: design the compressive stress, using the second design equation

[tex]f_{c} =f_{cr}-2.33*s + 500[/tex] -------equation 2

[tex]f_{c} =4780-2.33*525 + 500[/tex] = 4056.75 psi

Therefore, since both compressive stress is greater than 4000 psi, the concrete satisfies the requirements of ACI code for 4000 psi concrete


Related Questions

A four-lane divided multilane highway (two lanes in each direction) in rolling terrain has five access points per mile and 11-ft lanes with a 4-ft shoulder on the right side and 2-ft shoulder on the left. The peak-hour factor is 0.84 and the traffic stream has a heavy vehicle adjustment factor of 0.847. If the analysis flow rate is 1250 pc/h/ln, what is the peak-hour volume? 976 veh/h 1345 veh/h 1389 veh/h 1779 veh/h

Answers

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

Previously , you wrote a program named Hurricane that classified hurricanes into five categories using the Saffir-Simpson Hurricane Scale. Now, create a modified version named Hurricane Modularized that passes a user’s input wind speed to a method that returns the hurricane category.

Answers

Answer:

Answer  of the Java class explained below with appropriate comments

Explanation:

using System;

class MainClass {

 public static void Main (string[] args) {

//entering wind speed of the hurricane

   Console.WriteLine ("Enter wind speed in mph: ");

   int windSpeed = Convert.ToInt32(Console.ReadLine());

//checking for the category with the credited wind speed

   int category = GetCategory(windSpeed);

   Console.WriteLine ("Category: " + category);

 }

 public static int GetCategory(int wind) {

     //function for classifying the wind speeds

     if(wind >= 157) {

         

     return 5;

     }

     if(wind >= 130) {

         return 4;

     }

     if(wind >= 111) {

         return 3;

     }

     if(wind >= 96) {

         return 2 ;

     }

     return 1;

 }

}

What are nonexecuting statements that you can use to document or add notes to assist in the use of the program?

Answers

Answer:

Program comments

Write a function called pyramid(height) that acceptsa parameter ""height"". It then prints a pyramid of that height

Answers

Answer:

I am writing a function in C++                              

Explanation:

C++ program

#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

}

Given a Pane object appPane and a TextField object nameField, write a statement that adds nameField to the pane.

Answers

Answer:

appPane.getChildren().add(nameField);

Explanation:

The question is related to JavaFx which is a set of packages used for making interactive graphical user interface that contains graphical components for better user experience.

Pane is a JavaFx component which is used for adjusting the position and size of a graphical component for a given scene.

We can create a pane object by appPane = new Pane(); command.

TextField is a JavaFx object that is used for displaying a line of text. We have various functions available to set properties of TextField object.

We can create a TextField object by  nameField = new TextField(); command.

A pane can have multiple set of graphical components for various parts of an application. These are called children.

Hence we can write the following statement to add a TextField object nameField to a Pane object appPane.

appPane.getChildren().add(nameField);

For an atmospheric pressure of 101 kPa (abs) determine the heights of the fluid columns in
barometers containing one of the following liquids a) mercury, b) water, c) ethyl alcohol.

Calculate the heights including the effect of vapour pressure and compare the results with those obtained
neglecting vapour pressure. Do these results support the widespread use of a particular fluid for
barometers? Explain the reason.

Answers

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.

Write a program that adds the numbers 1 through 5 into the numbers ArrayList and then prints out the first element in the list.

Answers

Answer:

Answer code is given below along with comments to explain each step

Explanation:

Method 1:

// size of the array numbers to store 1 to 5 integers

int Size = 5;  

// here we have initialized our array numbers, the type of array is integers and size is 5, right now it is empty.

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// lets store values into the array numbers one by one

numbers.add(1)

numbers.add(2)

numbers.add(3)

numbers.add(4)

numbers.add(5)

// here we are finding the first element in the array numbers by get() function, the first element is at the 0th position

int firstElement = numbers.get(0);

// to print the first element in the array numbers

System.out.println(firstElement);  

Method 2:

int Size = 5;  

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// here we are using a for loop instead of adding manually one by one to store the values from 1 to 5

for (int i = 0; i < Size; i++)  

{

// adding the values 1 to 5 into array numbers

  numbers.add(i);  

}

//here we are finding the first element in the array numbers

int firstElement = numbers.get(0);  

// to print the first element in the array

System.out.println(firstElement);  

The article "Display of Health Risk Behaviors on MySpace by Adolescents"† 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).

Answers

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 net work output and the thermal efficiency for the Carnot and the simple ideal Rankine cycles with steam as the working fluid are to be calculated and compared. Steam enters the turbine in both cases at 5 MPa as a saturated vapor and the condenser pressure is 50 kPa. in the Rankine cycle, the condenser exit state is saturated liquid and in the Carnot cycle, the boiler inlet state is saturated liquid. Draw the T-s diagrams for both cycles.

Answers

Answer:

a) Rankine

Net work output = 719.1 KJ/kg

Thermal Eff = 0.294

a) Carnot

Net work output = 563.2 KJ/kg

Thermal Eff = 0.294

For T-s diagrams see attachments

Explanation:

Part a    Rankine Cycle

The obtained data from water property tables:

[tex]P_{L,sat liquid} = 50 KPa \\v_{1} = 0.00103m^3/kg\\\\ h_{1} = 340.54KJ/kg\\\\P_{H} = 5000KPa\\h_{2} = h_{1} + v_{1} *(P_{H} - P_{L} )\\h_{2} =350.54 + (0.00103)*(5000 - 50)\\\\h_{2} = 345.64KJ/kg\\\\P_{H,satsteam} = 5000KPa\\s_{3} = 5.9737KJ/kgK\\\\h_{3} = 2794.2KJ/kg\\\\s_{3} = s_{4} = 5.9739KJ/kgK\\P_{L}= 50KPa\\\\h_{4}= 2070KJ/kg\\\\[/tex]

Heat transferred from boiler

[tex]q_{b} = h_{3}-h_{2}\\q_{b}=2794.2-345.64\\\\q_{b} =2448.56KJ/kg\\\\[/tex]

Heat transferred from condenser

[tex]q_{c} = h_{4}-h_{1}\\q_{b}=2070-340.54\\\\q_{b} =1729.46KJ/kg\\\\[/tex]

Thermal Efficiency

[tex]u_{R} = 1- \frac{q_{c}}{q_{b}}\\\\u_{R} = 1 - \frac{1729.46}{2448.56}\\\\u_{R} =0.294[/tex]

Net work output

[tex]w_{R} = q_{b}-q_{c}\\w_{R} = 2448.56-1729.46\\\\w_{R}=719.1KJ/kg[/tex]

Part b    Carnot Cycle

The obtained data from water property tables:

[tex]P_{H,sat-steam} = 5000KPa\\T_{3} = 263.94 C\\s_{3} = 5.9737KJ/kgK\\\\h_{3} = 2794.2KJ/kg\\\\T_{2,sat-liquid} = T_{3} = 263.94C\\s_{2} = 2.920KJ/kgK\\\\h_{2} = 1150KJ/kg\\\\P_{L} = 50KPa\\s_{1}=s_{2} = 2.920KJ/kgK\\\\h_{1} = 989KJ/kg\\\\s_{3} = s_{4} = 5.9737KJ/kgK\\P_{L} = 50KPa\\\\h_{4} = 2070KJ/kg[/tex]

Heat transferred from boiler

[tex]q_{b} = h_{3}-h_{2}\\q_{b}=2794.2-1150\\\\q_{b} =1644.2KJ/kg\\\\[/tex]

Heat transferred from condenser

[tex]q_{c} = h_{4}-h_{1}\\q_{b}=2070-989\\\\q_{b} =1081KJ/kg\\\\[/tex]

Thermal Efficiency

[tex]u_{C} = 1- \frac{q_{c}}{q_{b}}\\\\u_{C} = 1 - \frac{1081}{1644.2}\\\\u_{C} =0.343[/tex]

Net work output

[tex]w_{C} = q_{b}-q_{c}\\w_{C} = 1644.2-1081\\\\w_{C}=563.2KJ/kg[/tex]

A scrubber on a coal-fired power plant is useful to remove

a. sulfur dioxide.
b. carbon dioxide.
c. particulate matter.

Answers

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:

A six- lane freeway ( three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hour factor is 0.80, and there are 8% large trucks and buses and 6% recreational vehicles in the traffic stream. One upgrade is 5% and 0.5 mi long. An analyst has determined that the freeway is operating at capacity on this upgrade during the peak hour. If the peak- hour traffic volume is 3900 vehicles, what value for the driver population factor was used?

Answers

Answer:

0.867

Explanation:

The driver population factor ([tex]f_{p}[/tex])can be estimated using the equation below:

[tex]f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}[/tex]

The value of the heavy vehicle factor ([tex]f_{HV}[/tex]) is determined below:

The values of the [tex]E_{T}[/tex] = 2 and [tex]E_{R}[/tex] = 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:

[tex]f_{HV}[/tex] = 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833

Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:

[tex]f_{p}[/tex] = 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867

Which of the following elements of the CIA triad refers to maintaining and assuring the accuracy of data over its life-cycle?

Confidentiality

Integrity

Availability

Authentication

Answers

Answer:

Integrity: involves maintaining and assuring the accuracy of data over its life-cycle

Explanation:

Confidentiality: This is a CIA triad designed to prevent sensitive information from reaching the wrong people, while making sure that the right people have access to it.

Integrity: This is a CIA triad that involves maintaining the consistency, accuracy, and trustworthiness of data over its entire life cycle.

Availability: This is a CIA triad that involves hardware repairs and maintaining a correctly functioning operating system environment that is free of software conflicts.

Authentication:This is a security control that is used to protect the system with regard to the CIA properties.

Liquid toluene is flowing through a pipe at a rate of 175 m 3 / h 175m3/h . What is the mass flow rate of this stream in kg/min? What is the molar flow rate in mol/s? In fact, the answer to Part (a) is only an approximation that is almost certain to be slightly in error. What did you have to assume to obtain the answer?

Answers

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.

A thermistor is a temperature‐sensing element composed of a semiconductor material, which exhibits a large change in resistance proportional to a small change in temperature. A particular thermistor has a resistance of 5 kΩ at 25°C. Its resistance is 340 Ω at 100°C. Assuming a straight‐line relationship between these two values, at what temperature will the thermistor's resistance equal 1 kΩ?

Answers

The temperature at which the resistance is [tex]1 k\Omega[/tex] is [tex]89.4^{\circ}C[/tex]

Explanation:

For the thermistor in this problem, the relationship between temperature and resistance is linear.

We have:

[tex]R_1 = 5000 \Omega[/tex] when the temperature is [tex]T=25^{\circ}C[/tex]

[tex]R_2=340 \Omega[/tex] when the temperature is [tex]T=100^{\circ}C[/tex]

Assuming a straight-line relationship, we can find the slope of the line:

[tex]m=\frac{R_2-R_1}{T_2-T_1}=\frac{340-5000}{100-25}=-62.1 \Omega/^{\circ}C[/tex]

Now that we know the slope, we can extrapolate the temperature when the resistance is

[tex]R_3 = 1 k\Omega = 1000 \Omega[/tex]

In fact, we can write:

[tex]m=\frac{R_3-R_2}{T_3-T_2}[/tex]

And solving for [tex]T_3[/tex],

[tex]m(T_3-T_2)=R_3-R_2\\T_3 = T_2 +\frac{R_3-R_2}{m}=100 + \frac{1000-340}{-62.1}=89.4^{\circ}C[/tex]

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Final answer:

The temperature at which the thermistor's resistance equals 1 kΩ is approximately 11.88 °C.

Explanation:

The relationship between the resistance of a thermistor and temperature can be modeled by a straight line equation, which is given by:

R = Ro(1 + αΔT)

where R is the resistance at a given temperature, Ro is the resistance at a reference temperature, α is the temperature coefficient of resistance, and ΔT is the change in temperature from the reference temperature.

To find the temperature at which the thermistor's resistance equals 1 kΩ, we can rearrange the equation:

ΔT = (R/Ro - 1) / α

Plugging in the given values:

Ro = 5 kΩ, R = 1 kΩ, α = (340 Ω - 5 kΩ) / (100 °C - 25 °C)

ΔT = (1 kΩ / 5 kΩ - 1) / [(340 Ω - 5 kΩ) / (100 °C - 25 °C)]

Simplifying the equation, we find that ΔT ≈ -14.12 °C.

Therefore, the temperature at which the thermistor's resistance equals 1 kΩ is approximately 11.88 °C.

Determine the tension developed in cord DE required for equilibrium of the lamp. Express your answer to three significant figures and include the appropriate units.

Answers

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)

A wastewater treatment plant discharges 1.0 m3/s of effluent having an ultimate BOD of 40.0 mg/ L into a stream flowingat 10.0 m3/s. Just upstream from the discharge point, the stream has an ultimate BOD of 3.0 mg/L The deoxygenationconstant kd is estimated at 0.22/day.(a) Assuming complete and instantaneous mixing. find the ultimate BOD of the mixture of waste and river just downstreamfrom the outfall.(b) Assuming a constant cross-sectional area for the stream equal to 55 m2, what ultimate BOD would you expect to find ata point 10.000 m downstream?

Answers

Answer:

a) 6.4  mg/l

b) 5.6 mg/l

Explanation:

Given data:

effluent Discharge Q_w = 1.0 m^3.s

Ultimate BOD L_w = 40 mg/l

Discharge of stream Q_r = 10 m^3.s

Stream ultimate BOD L_r = 3  mg/l

a) Ultimate BOD of mixture[tex] = \frac{Q_w l_w + Q_r L_r}{Q_w + Q_r}[/tex]

                                         [tex] = \frac{1*40 + 10*3}{10 +1} = 6.4 mg/l[/tex]

b) utlimate BOD at 10,000 m downstream

[tex]t =\frac{distance}{speed} = \frac{10,000}{\frac{Q_r +Q+w}{55}} \times \frac{hr}{3600} \times  \frac{day}{24 hr}[/tex]

putting [tex]Q_r + Q_w = 1+ 10 = 11 m^3/s[/tex]

t = 0.578  days

we know

[tex]L_t = L_o e^{-kt}[/tex]

[tex]L_t = 6.4 \times e^{-0.22 \times 0.578}[/tex]

[tex]L_t = 5.6 mg/l[/tex]

Write a program with total change amount as an integer input, and output the change using the fewest coins, one coin type per line. The coin types are Dollars, Quarters, Dimes, Nickels, and Pennies. Use singular and plural coin names as appropriate, like 1 Penny vs. 2 Pennies.

Answers

A program with total change amount as an integer input, and output the change using the fewest coins, one coin type per line is in the explanation part below.

Python program that outputs the change using the fewest coins after receiving the entire amount of change as an integer input:

def calculate_change(change_amount):

   coin_types = {

       "Dollar": 100,

       "Quarter": 25,

       "Dime": 10,

       "Nickel": 5,

       "Penny": 1

   }

   print("Change breakdown:")

   for coin_name, coin_value in coin_types.items():

       if change_amount >= coin_value:

           num_coins = change_amount // coin_value

           change_amount -= num_coins * coin_value

           coin_name_plural = coin_name + "s" if num_coins > 1 else coin_name

           print(f"{num_coins} {coin_name_plural}")

# Input the total change amount

total_change = int(input("Enter total change amount: "))

calculate_change(total_change)

Thus, this is the Python program asked.

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How much energy in joules does a 12 volt battery in your car use to move 2.5 coulombs through the electrical circuit?

Answers

Answer: 30joules

Explanation

voltage of battery =12volts

2.5coulombs is the same as 2.5ampere-seconds

Energy required =12V*2.5A-S

=30watts-seconds

Hence the required energy is equal to 30joules.

A signal whose timing is completely influenced by the traffic volumes, when detected, on all of the approaches operates in the following mode: a. Pretimed b. Semi-actuated c. Fully-actuated

Answers

Answer: c. Fully actuated

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

The loss of power a signal suffers as it travels from the transmitting computer to a receivingcomputer is: a.white noiseb.spikingc.attenuation d.intermodulation noisee.echo

Answers

Answer:

c.attenuation

Explanation:

In telecommunication, it is called attenuation of a signal, be it acoustic, electrical or optical, to the loss of power suffered by it when passing through any transmission medium.

Attenuation is usually not expressed as a difference in powers but in logarithmic units such as decibels, which are more comfortable to use when calculating.

The attenuation is expressed in decibels (db) and the power is measured with this formula:

α = 10 * log (P1 / P2)

where P is the power both initial and final.

Which of the following has nothing to do with insulating glass? Group of answer choices

a. triple glazing
b. double glazing
c. low emissivity coatings
d. low conductivity
e. gas between sheets of glass
f. tempered glass

Answers

Answer:

Tempered glass

Explanation:

Tempered or toughened glass is a type of safety glass processed by controlled

thermal or chemical treatments to increase its strength compared with normal glass. Tempering puts the outer surfaces into compression and the interior into tension . Such stresses cause the glass, when broken, to crumble into small granular chunks instead of splintering into jagged shards as plate glass (a.k.a. annealed glass) does. The granular chunks are less likely to cause injury.

Answer: tempered glass

The characteristics listed in the options are the characteristics of an insulating glass, a type of glass with double or more panes used to reduce heat loss within a building.

Explanation:

A tempered glass is also known as a toughened glass. It is a type of glass that is produced through controlled thermal or chemical treatment which helps to increase its quality in terms of toughness. A tempered glass is a safety glass and as a result of this, its surface is four times toughened than that of a normal glass.

A certain lead-acid storage battery has a mass of 33 kg . Starting from a fully charged state, it can supply 6 A for 20 hours with a terminal voltage of 24 V before it is totally discharged.

a. If the energy stored in the fully charged battery is used to lift the battery with 100-percent efficiency, what height is attained? Assume that the acceleration due to gravity is 9.8 m/s2 and is constant with height.

b. If the stored energy is used to accelerate the battery with 100 percent efficiency, what velocity is attained?

c. Gasoline contains about rho1 = 4.5×107 J/kg. Compare this with the energy content per unit mass for the fully charged battery (rho2).

ans: rho2/ rho1=?

Answers

Answer: attached below. rate brainliest if correct please

Consider as a system the gas in a vertical cylinder; the cylinder is fitted with a piston onwhich a number of small weights are placed. The initial pressure is 200 kPa, and the initial volume ofthe gas is 0.04 m^3. Let a bunsen burner be placed under the cylinder, and let thevolume of the gas increase 0.1m^3 while the pressure remainsconstant.
A. Calculate the work done by the system during this process.

Answers

Answer:

Work done = 12,000J = 12KJ

Explanation:

This process is an Isobaric one i.e a process at constant pressure

P = 200KPa = 200 x 1000 Pa = 200,000PaVi = 0.04m3Vf = 0.1m3

From the formula for Workdone = Integral (PdV)

Work done = P ( Vf - Vi)

Plugging the values in the equation,

Work done = 200,000 ( 0.1 - 0.04)

Work done = 12,000J = 12KJ

Give two circumstances in which in situ treatment would be better, and two in which ex situ treatment would be better. Both using chemical oxidation as the remediation technique.

Answers

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;

a large volume of soil is to be treated at oncea permeable sandy soil (un-compacted) is to be treated

ex situ treatment would be better would be better if;

a wider range of contaminants and soil types are to be treatedthere is more certainty about the uniformity of treatment because of the ability to homogenize, screen, and continuously mix the soil.

An automotive Battery is rated at120 A-h. This means that under certain test conditions, it canoutput 1 A at 12 V for 120 hours.

(a) How much total energy is storedin battery?

(b) If the headlights are left on overnight (8h),how much energy will still be stored in the battery in the morning(Assume a 150 w total power rating for both headlightstogether)

Answers

Answer:

Part A:

In W-h:

Energy Stored=1440 W-h

In Joules:

[tex]Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ[/tex]

Part B:

In W-h:

Energy left=240 W-h

In Joules:

Energy left= 8.64*10^5 J

Explanation:

Part A:

We are given rating 120A-h and voltage 12 V

Energy Stored= Rating*Voltage              (Gives us units W-h)

Energy Stored=120A-h*12V

Energy Stored=1440 W-h

Converting it into joules (watt=joules/sec)

Energy Stored=[tex]1440 Joules * \frac{3600sec}{h}h[/tex]

Energy Stored=5184000 Joules

[tex]Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ[/tex]

Part B:

Energy used by lights for 8h=150*8

Energy used by lights for 8h=1200W-h

Energy left= Energy Stored(Calculated above)- Energy used by lights for 8h

Energy left=1440-1200

Energy left=240 W-h

Energy left=[tex]240 Joules * \frac{3600sec}{h}h[/tex]

Energy left=864000 Joules

Energy left= 8.64*10^5 J

Final answer:

The total energy stored in the battery is 1440 Wh. If the headlights are left on overnight (8h), the energy remaining in the battery in the morning is 1320 Wh.

Explanation:

(a) The total energy stored in the battery can be calculated by multiplying the battery's capacity (120 A-h) by the voltage (12 V). So, the total energy stored in the battery is 1440 Wh.

(b) To calculate the energy consumed by the headlights in 8 hours, we need to find the power (P) using the formula P = V × I, where V is the voltage (12 V) and I is the current (the total power rating divided by the voltage). Then, we can find the energy consumed by multiplying the power by the time (8 hours). We can subtract this energy consumption from the total energy stored in the battery to find the energy remaining in the morning. So, the energy remaining in the battery is 1440 Wh - 12 kWh = 1320 Wh.

Learn more about Energy stored in an automotive battery here:

https://brainly.com/question/34222184

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The viscosity of a fluid is to be measured by an viscometer constructed of two 5 ft long concentric cylinders. The inner diameter of the outer cylinder is 6 in, and the gap between the two cylinders is 0.035 in. The outer cylinder is rotated at 250 rpm, and the torque is measured to be 1.2lbf *ft.
A) Determine the viscosity of the fluid.

Answers

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]

The pressure in a water line is 1500 kPa. What is the line pressure in (a) lb/ft2units and (b) lbf/in2(psi) units?

Answers

Answer:

Part A:

[tex]1500\ KPa= 31328.145 \frac{lb}{ft^2}[/tex]

Part B:

[tex]1500 KPa=217.55656 \frac{lb}{in^2}[/tex](Psi)

Explanation:

Part A:

Line Pressure is 1500 KPa

We need a conversion factor which converts KPa to lb/ft^2.

[tex]20.88543 \frac{lb}{ft^2}= 1\ KPa[/tex]

In order to convert 1500 KPa to lb/ft^2, we proceed as:

[tex]1\ KPa=20.88543 \frac{lb}{ft^2} \\1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa}\\1500\ KPa= 31328.145 \frac{lb}{ft^2}[/tex]

1500 KPa is 31328.145 lb/ft^2

Part B:

We will use the same procedure we did in Part A:

1 ft= 12 in

[tex](1\ ft)^2=(12\ in)^2\\1 ft^2=144 in^2[/tex]

Converting [tex]1500 KPa\ into\ \frac{lb}{in^2}[/tex]

[tex]1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa} * \frac{ft^2}{144\ in^2}[/tex]

[tex]1500 KPa=217.55656 \frac{lb}{in^2}[/tex](Psi)

1500 KPa is 217.55656 lb/in^2 (psi)

A certain battery has terminals labeled a and b. The battery voltage is vab = 12 V. a) To increase the chemical energy stored in the battery by 900J, how much charge must move through the battery? b) Should electrons move from a to b or from b to a?

Answers

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

Create a conditional expression that evaluates to string "negative" if userVal is less than 0, and "non-negative" otherwise. Ex: If userVal is -9, output is: -9 is negative.

Answers

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".

Chlorate (ClO3 − ) is a toxic anion in polluted water from the use of high energy materials, herbicide, and drinking water disinfection with chlorine chemicals. In a reaction for the catalytic reduction of ClO3 − into the non-toxic chloride (Cl− ), the reaction rate constant was measured as 9.98 hr−1 , then how many minutes does the reaction take to reduce 90% of the chlorate in the water?

Answers

Answer:

Time taken = 5.41mins

Explanation:

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

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