During an actual run with the calorimeter, there are two important criteria which must be satisfied by a correctly positioned thermometer. What are these two criteria?

Answers

Answer 1

Answer:

1. the bulb of the thermometer must be in the center of the lower calorimeter, about one inch from the bottom.

2.thermometer must face you to read

Explanation:

During an actual run with the calorimeter, there are two important criteria which must be satisfied by a correctly positioned thermometer. These two criteria include;

1. the bulb of the thermometer must be in the center of the lower calorimeter, about one inch from the bottom.

2. thermometer must face you to read

Answer 2

The thermometer must be fully immersed without touching the calorimeter’s sides or bottom and positioned to minimize heat exchange with the external environment to ensure accurate temperature readings.

When using a calorimeter, it is crucial to ensure that the thermometer is correctly positioned to obtain accurate measurements. The two important criteria for correctly positioning the thermometer in a calorimeter are:

1. Thermal Equilibrium:

  - The thermometer must be fully immersed in the substance being measured (typically a liquid) without touching the sides or the bottom of the calorimeter. This ensures that the thermometer accurately reflects the temperature of the substance rather than the container.

  - The immersion should be deep enough so that the thermometer's bulb is surrounded by the substance, allowing it to reach thermal equilibrium with the substance. This helps in obtaining a consistent and accurate temperature reading.

2. Minimize Heat Exchange with the Environment:

  - The thermometer should be positioned in a way that minimizes heat exchange with the external environment. The calorimeter is designed to be an insulated system, so the thermometer should not introduce significant thermal disturbance.

  - The top of the calorimeter should be closed or covered as much as possible to prevent heat loss or gain from the surroundings, which could affect the temperature reading. The thermometer should pass through a small hole or a sealed opening to maintain the insulation integrity of the calorimeter.

By adhering to these criteria, the thermometer can provide precise and reliable temperature measurements, which are essential for accurate calorimetric calculations.


Related Questions

If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the conductor.

Answers

Answer:

The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²

Explanation:

The electric field intensity due to a thin conducting sheet is given by the following formula:

Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)

From this formula:

Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)

We have the following data:

Electric Field Intensity = 1.5 N/C

Permittivity of free space = 8.85 x 10^-12 C²/N.m²

Therefore,

Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)

Surface Charge Density = 26.55 x 10^-12 C/m²

Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².

The surface charge density on the conductor is; σ = 13.275 × 10⁻¹² C/m²

What is the surface charge density?

The formula for surface charge density on a conductor in an electric field just outside the surface of conductor is;

σ = E * ϵ₀

​where;

E is electric field = 1.5 N/C

ϵ₀ is permittivity of space = 8.85 × 10⁻¹² C²/N.m²

Thus;

σ = 1.5 * 8.85 × 10⁻¹²

σ = 13.275 × 10⁻¹² C/m²

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Assume we have already defined a variable of type String called password with the following line of code: password' can have any String value String password ???"; However, because of increased security mandated by UCSD's technical support team, we need to verify that passwo rd is secure enough. Specifically, assume that a given password must have all of the following properties to be considered "secure": It must be at least 7 characters long .It must have characters from at least 3 of the following 4 categories: Uppercase (A-Z), Lowercase (a-z), Digits (0-9), and Symbols TASK: If password is secure, print "secure", otherwise, print "insecure" HINT: You can assume that any char that is not a letter (A-Z, a-z) and is not a digit (0-9) is a symbol. EXAMPLE: "UCSDcse11" would be valid because it is at least 7 characters long (it's 9 characters long), it has at least one uppercase letter (U', 'C', 'S', and 'D'), it has at least one lowercase letter (c', 's', and 'e'), and it has at least one digit (1' and '1') EXAMPLE: "UCSanDiego" would be invalid because, while it is 10 characters long, it only has uppercase and lowercase letters, so it only has characters from 2 of the 4 categories listed Sample Input: UCSDcse11 Sample Output: secure

Answers

Answer:

The Java code is given below with appropriate comments for better understanding

Explanation:

import java.util.Scanner;

public class ValidatePassword {

  public static void main(String[] args) {

      Scanner input = new Scanner(System.in);

          System.out.print("Enter a password: ");

          String password = input.nextLine();

          int count = chkPswd(password);

          if (count >= 3)

              System.out.println("Secure");

          else

              System.out.println("Not Secure");

  }

  public static int chkPswd(String pwd) {

      int count = 0;

      if (checkForSpecial(pwd))

          count++;

      if (checkForUpperCasae(pwd))

          count++;

      if (checkForLowerCasae(pwd))

          count++;

      if (checkForDigit(pwd))

          count++;

      return count;

  }

  // checks if password has 8 characters

  public static boolean checkCharCount(String pwd) {

      return pwd.length() >= 7;

  }

  // checks if password has checkForUpperCasae

  public static boolean checkForUpperCasae(String pwd) {

      for (int i = 0; i < pwd.length(); i++)

          if (Character.isLetter(pwd.charAt(i)) && Character.isUpperCase(pwd.charAt(i)))

              return true;

      return false;

  }

  public static boolean checkForLowerCasae(String pwd) {

      for (int i = 0; i < pwd.length(); i++)

          if (Character.isLetter(pwd.charAt(i)) && Character.isLowerCase(pwd.charAt(i)))

              return true;

      return false;

  }

  // checks if password contains digit

  public static boolean checkForDigit(String pwd) {

      for (int i = 0; i < pwd.length(); i++)

          if (Character.isDigit(pwd.charAt(i)))

              return true;

      return false;

  }

  // checks if password has special char

  public static boolean checkForSpecial(String pwd) {

      String spl = "!@#$%^&*(";

      for (int i = 0; i < pwd.length(); i++)

          if (spl.contains(pwd.charAt(i) + ""))

              return true;

      return false;

  }

}

My Notes How many grams of perchloric acid, HClO4, are contained in 39.1 g of 74.9 wt% aqueous perchloric acid?
How many grams of water are in the same solution?

Answers

Answer:

a)29.9 b) 9.81

Explanation:

Wt% = mass of solute / mass of solvent × 100

0.749 = mass of solute / mass of solvent

a) Mass of perchloric acid = 0.749 × 39.1 = 29.29

b) Mass of water = 39.1 - 29.29 = 9.81

The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 13.2 ft/sft/s at point A and 17.6 ft/sft/s at point C. The cart takes 3.00 ss to go from point A to point C, and the cart takes 1.90 ss to go from point B to point C. What is the car's speed at point B?

Answers

Answer:

15.99 ft/s

Explanation:

From Newton's equation of motion, we have

v = u + at

v = Final speed

u = initial speed

a = acceleration

t = time

now

for the points A and C

v = 17.6 ft/s

u = 13.2 ft/s

t = 3 s

thus,

17.6 = 13.2 + a(3)

or

3a = 17.6 - 13.2

3a = 4.4

or

a = 1.467 m/s²

Thus,

For Points A and B

v = speed at B i.e v'

u = 13.2 ft/s

a = 1.467 ft/s²

t = 1.90 s

therefore,

v' = 13.2 + (1.467 × 1.90 )

v' = 13.2  + 2.7867

v' = 15.9867 ≈ 15.99 ft/s

(1 point) Consider the initial value problem 2????y′=4y, y(−1)=−2. Find the value of the constant ???? and the exponent ???? so that y=???????????? is the solution of this initial value problem. y= help (formulas) Determine the largest interval of the form ????<????<???? on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution. help (inequalities) What is the actual interval of existence for the solution (from part a)? help (inequalities)

Answers

Answer:

Here is the missing part of the question ;

find the value of the constant C and the exponent r so that y = Ctr is the solution of this initial value problem. Determine the largest interval of the form a<t<b

Explanation:

The step by step explanation is as given in the attachment. You will notice I used β as the exponential in place of r.

The method of breaking digital messages into small fixed bundles, transmitting them along different communication paths, and reassembling them at their destination is called:

Answers

Answer:

Packet switching

Explanation:

Packet switching - it is referred to as the transmission process which involved sending of packets through communication path and last reassembling them again.

The sending of packets is done by broken the digital message into required pieces for efficient transfer.  

in this process, every individual parcel is sent separately.

A cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.

Answers

Answer:

1.505

Explanation:

cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.

stress is force per unit area

stress=P/A

A = πd^2/4.

uncertainty of axial force P= +/-.11

s=+/-.20, strength

d=+/-.04 diameter

fail load/max allowed

minimum design=fail load/max allowed

minimum design =s/(P/A)

sA/P

A=([tex]\pi[/tex].96d^2)/4, so Amin=

[tex]0.96^{2}[/tex] (because the diameter  at minimum is (1-0.04=0.96)

minimum design=Pmax/(sminxAmin)

1.11/(.80*.96^2)=

1.505

The net potential energy between two adjacent ions, EN, may be represented by EN = -A/r + B/rn Where A, B, and n are constants whose values depend on the particular ionic system. Calculate the bonding energy E0 in terms of the parameters A, B, and n using the following procedure: 1. Differentiate EN with respect to r, and then set the resulting expression equal to zero, since the curve of EN versus r is a minimum at E0. 2. Solve for r in terms of A, B, and n, which yields r0, the equilibrium interionic spacing. 3. Determine the expression for E0 by substitution of r0 into the above equation for EN. What is the equation that represents the expression for E0?

Answers

The  equation that represents the expression for the bonding energy [tex](E_0\))[/tex] in terms of the parameters A, B, and n is: [tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex].

Step 1: Differentiate EN with Respect to r

Given:

[tex]\[ E_N = -\frac{A}{r} + \frac{B}{r^n} \][/tex]

Differentiate [tex]\(E_N\)[/tex] with respect to r:

[tex]\[ \frac{dE_N}{dr} = \frac{A}{r^2} - \frac{nB}{r^{n+1}} \][/tex]

Set the derivative equal to zero to find the minimum (equilibrium) point:

[tex]\[ \frac{A}{r^2} - \frac{nB}{r^{n+1}} = 0 \][/tex]

Solve for r as

[tex]\[ \frac{A}{r^2} = \frac{nB}{r^{n+1}} \][/tex]

[tex]\[ r^n = \frac{nB}{A} \][/tex]

[tex]\[ r = \left(\frac{nB}{A}\right)^{\frac{1}{n}} \][/tex]

Step 2: Solve for r in terms of A, B, and n

The equilibrium interionic spacing [tex]\(r_0\)[/tex] is the value of r at the minimum point,

[tex]\[ r_0 = \left(\frac{nB}{A}\right)^{\frac{1}{n}} \][/tex]

Step 3: Determine the Expression for E0

Substitute [tex]\(r_0\)[/tex] into the expression for [tex]\(E_N\):[/tex]

[tex]\[ E_0 = -\frac{A}{r_0} + \frac{B}{r_0^n} \][/tex]

[tex]\[ E_0 = -\frac{A}{\left(\frac{nB}{A}\right)^{\frac{1}{n}}} + \frac{B}{\left(\frac{nB}{A}\right)^{\frac{n}{n}}} \][/tex]

Simplify:

[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{B}{(nB)^{\frac{n}{n}}} \][/tex]

[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{B}{nB} \][/tex]

[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex]

Combine the terms:

[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex]

So, the equation is [tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex].

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Explain the differences among sand, silt, and clay, both in their physical characteristics and their behavior in relation to building foundations.

Answers

Answer:PHYSICALLY- sand is spherical in shape with large particles up to 0.18g

Silt also contains more of spherical particles with weights around 0.0029g.

Clay contain plate like particles,which are flat or layered. It's particles are generally lower that silt and sand below 0.0029g.

BEHAVIORS Sand particles are coarse and very porous,water can easily penetrate with no particles cohesion,does not retain water, difficult to expand,it is IDEAL FOR BUILDINGS.

Silt particles are also porous with some coarseness and no particles cohesion. retains water and can expand. NOT IDEAL FOR BUILDINGS.

Clay contain particles that are not coarse,they have high cohesiveness,they are not porous as it is difficult for water to penetrate. NOT IDEAL FOR BUILDINGS.

Explanation: Sand particles are coarse,very porous,contains spherical particles with large particles sizes and IDEAL FOR BUILDINGS

Silt particles are also porous, with some coarseness and with little or no particles cohesion, NOT IDEAL FOR BUILDINGS.

Clay soils are not porous water can not easily flow thought it,it is hard when dry and soft when wet. IT IS NOT A GOOD OPTION FOR BUILDINGS.

Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of the time and the unit cost of electricity is $0.09/kWh, the electricity cost of this refrigerator per month (30 days) is

A. $3.56
B. $5.18
C. $8.54
D. $9.28
E. $20.74

Answers

Answer:

B. $5.18

Explanation:

Cost of electricity per kWh = $0.09

Power consumption of refrigerator = 320W = 320/1000 = 0.32kW

In a month (30 days) the refrigerator works 1/4 × 30 days = 7.5 days = 7.5 × 24 hours = 180 hours

Energy consumed in 180 hours = 0.32kW × 180h = 57.6kWh

Cost of electricity of 57.6kWh energy consumed by the refrigerator = 57.6 × $0.09 = $5.18

Final answer:

The electricity cost of the refrigerator per month is approximately $0.65. The correct answer is none of the given options.

Explanation:

To calculate the electricity cost of the refrigerator per month, we first need to find out how many kWh it consumes when it is running for one hour. We can do this by converting the power consumption of the refrigerator from watts to kilowatts:

320 W = 0.32 kW

Since the refrigerator runs only one quarter of the time, we can calculate the kWh consumed per month as follows:

0.32 kW * 1/4 * 24 hours/day * 30 days/month = 7.2 kWh/month

Now, we can calculate the cost of electricity using the unit cost of $0.09/kWh:

7.2 kWh/month * $0.09/kWh = $0.648/month

Therefore, the electricity cost of the refrigerator per month is approximately $0.65.

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Check my work Check My Work button is now disabledItem 16Item 16 3 points As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter changes from 1 cm to 3 cm. Calculate the amount of work produced by this bubble, in kJ, if the surface tension of ammonia is 0.07 N/m.

Answers

Answer:

W = 1.7593 * 10 ^ (-7) KJ

Explanation:

The work done by the bubble is given:

[tex]W = sigma*\int\limits^2_1 {} \, dA \\\\W = sigma*( {A_{2} - A_{1} } ) \\\\A = pi*D^2\\\\W = sigma*pi*(D^2_{2} - D^2_{1})\\\\W = 0.07 * pi * (0.03^2 - 0.01^2)*10^(-3)\\\\W = 1.7593 *10^(-7) KJ[/tex]

Answer: W = 1.7593 * 10 ^ (-7) KJ

The gage pressure in a liquid at a depth of 3 m is read to be 48 kPa. Determine the gage pressure in the same liquid at a depth of 9 m.
The gage pressure in the same liquid at a depth of 9 m is__________ kPa.

Answers

Answer: 144kpa

Explanation:

pressure density =p

The Gage pressure (P1)=48kpa

The height (h1)=3m

The Gage pressure (P2)=?

The height (h2)=9m

P=p * g * h

P1/P2=p * g * h1/p * g * h2

Cancelling out similar terms:

Therefore, P1/P2=h1/h2

P2=P1*h2/h1

Hence, P2=48*9/3=144kpa.

In a residential heat pump, Refrigerant-134a enters the condenser at 800 kPa and 50oC at a rate of 0.022 kg/s and leaves at 750 kPa subcooled by 3oC. The refrigerant enters the compressor at 200 kPa superheated by 4oC. Determine (a) the isentropic efficiency of the compressor, (b) the rate of heat supplied to the heated room, and (c) the COP of the heat pump. Also, (d) determine the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the pressure limits of 200 kPa and 800 kPa.

Answers

Answer:

a) 0.813

b) 4.38 KW

c) COP = 5.13

d) 3.91 KW , COP = 6.17

Explanation:

Data obtained A-13 tables for R-134a:

[tex]h_{1} = 247.88 \frac{KJ}{kg} \\s_{1} = 0.9575 \frac{KJ}{kgK}\\h_{2s} = 279.45 \frac{KJ}{kg}\\h_{2} = 286.71 \frac{KJ}{kg}\\h_{3} = 87.83 \frac{KJ}{kg}[/tex]

The isentropic efficiency of the compressor is determined by :

[tex]n_{C} = \frac{h_{2s} - h_{1} }{h_{2} - h_{1} }\\= \frac{279.45 - 247.88 }{286.71 - 247.88}\\= 0.813[/tex]

The rate of heat supplied to the room is determined by heat balance:

[tex]Q = m(flow) * (h_{2} -h_{3})\\= (0.022)*(286.71 - 87.83)\\\\= 4.38KW[/tex]

Calculating COP

[tex]COP = \frac{Q_{H} }{W} \\COP = \frac{Q_{H} }{m(flow) * (h_{2} - h_{1}) }\\\\COP = \frac{4.38}{(0.022)*(286.71-246.88)}\\\\COP = 5.13[/tex]

Part D

Data Obtained:

[tex]h_{1} = 244.5 \frac{KJ}{kg} \\s_{1} = 0.93788 \frac{KJ}{kgK}\\h_{2} = 273.31 \frac{KJ}{kg}\\h_{3} = 95.48 \frac{KJ}{kg}[/tex]

The rate of heat supplied to the room is determined by heat balance:

[tex]Q = m(flow) * (h_{2} -h_{3})\\= (0.022)*(273.31 - 95.48)\\\\= 3.91KW[/tex]

Calculating COP

[tex]COP = \frac{Q_{H} }{W} \\COP = \frac{Q_{H} }{m(flow) * (h_{2} - h_{1}) }\\\\COP = \frac{4.38}{(0.022)*(273.31-244.5)}\\\\COP = 6.17[/tex]

In each case indicate whether the quantity in question increased, decreased or stayed the same when the string length is increased. Assume that the tension is unchanged. The function generator is kept at the same frequency, and the string is in resonance in all cases. Part A Number of antinodes ___ Number of antinodes ___ increased. decreased. stayed the same. Request Answer Part B Wavelength ___ Wavelength ___ increased. decreased. stayed the same. Request Answer Part C Fundamental frequency ___ Fundamental frequency ___ increased. decreased. stayed the same.

Answers

Answer:

Answer: No of anti-nodes increases

Answer: wavelength remains same.

Answer: fundamental frequency decreases

Explanation:

a)

The number of nodes (n) would have (n-1) anti-nodes.

The relation of Length of string with n is given below:

[tex]L = \frac{n*lambda}{2}[/tex]

Hence, n and L are directly proportional so as string length increases number of nodes and anti-nodes also increases.

Answer: No of anti-nodes increases

b)

Wavelength is dependent on the frequency:

[tex]lambda = \frac{v}{f}[/tex]

The speed v of the string remains same through-out and frequency of generator is unchanged!

Hence according to above relationship lambda is unchanged.

Answer: wavelength remains same.

C

Fundamental frequency equates to 1st harmonic that 2 nodes and 1 anti-node. Wavelength is = 2*L

Hence, if L increases wavelength increases

Using relation in part b

As wavelength increases fundamental frequency decreases

Answer: fundamental frequency decreases

Design a sequential circuit with two D flip-flops A and B, and one input x_in.



(a)When xin = 0, the state of the circuit remains the same. When x_in = 1, the circuit goes through the state transitions from 00 to 01, to 11, to 10, back to 00, and repeats.



(b)When xin = 0, the state of the circuit remains the same. When x_in =1, the circuit goes through the state transitions from 00 to 11, to 01, to 10, back to 00, and repeats.

Answers

Answer:

View Image

Explanation:

The question is basically asking you to build a 2-bit asynchronous counter.

What the counter does is it increase it's value by 01₂ every clock pulse. So at 0₂, nothing happens, but at 1₂ it'll count up by 1. It then reset to 00₂ when it overflows.

The design for it is pretty much universal so I kinda did this from memory.

a.) A count-up counter (from 00-11) is simply made by connecting Q' to D, and the output of the previous DFF to the clock of the next one.

b.) A count-down counter (from 11-00) is simply made by using the same circuit as the count-up counter, but you connect Q' to the clock instead of Q.

Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) Cylindrical wall (b) Plane shell (c) Spherical shell (d) All of them (e) None of them

Answers

Answer:

Plane shell which is option b

Explanation:

The temperature in the case of a steady one-dimensional heat conduction through a plane wall is always a linear function of the thickness. for steady state dT/dt = 0

in such case, the temperature gradient dT/dx, the thermal conductivity are all linear function of x.

For a plane wall, the inner temperature is always less than the outside temperature.

A sample of coarse aggregate has an oven dry weight of 1034.0 g and a moisture content of 4.0 %. The saturated surface dry weight is 1048.9g and the weight of the aggregate in water is 675.6 g. Determine using phase volume relationships: a) Apparent Specific Gravity (GA) b) Bulk Specific Gravity (GB) c) Bulk Specific Gravity SSD (GB (SSD)) d) Absorption, % e) Bulk Volume

Answers

Answer:

Apparent Specific Gravity = 2.88

bulk specific gravity = 2.76

Bulk Specific Gravity SSD  = 2.80

absorption = 1.44%

bulk volume  = 373.3

Explanation:

given data

oven dry weight A  = 1034.0 g

moisture content = 4.0 %

saturated surface dry weight B = 1048.9 g

weight of the aggregate in water C = 675.6 g

solution

we get here Apparent Specific Gravity that is express as

Apparent Specific Gravity = [tex]\frac{A}{A-C}[/tex]   ..........1

put here value

Apparent Specific Gravity = [tex]\frac{1034}{1034-675.6}[/tex]

Apparent Specific Gravity = 2.88

and

now we get bulk specific gravity that is

bulk specific gravity = [tex]\frac{A}{B-C}[/tex]   ...................2

put here value

bulk specific gravity = [tex]\frac{1034}{1048.9-675.6}[/tex]

bulk specific gravity = 2.76

and

now we get Bulk Specific Gravity SSD

Bulk Specific Gravity SSD = [tex]\frac{B}{B-C}[/tex]    ............3

Bulk Specific Gravity SSD = [tex]\frac{1048.9}{1048.9-675.6}[/tex]

Bulk Specific Gravity SSD  = 2.80

and

now absorption will be here as

absorption = [tex]\frac{B-A}{A}[/tex] × 100%    ................4

absorption = [tex]\frac{1048.9-1034}{1034}[/tex] × 100%

absorption = 1.44%

and

last we get bulk volume that is

bulk volume = [tex]\frac{weight\ displce\ water}{density\ water }[/tex]

bulk volume = [tex]\frac{1048.9-675.6}{1}[/tex]

bulk volume  = 373.3

One AA battery in a flashlight stores 9400 J. The three LED flashlight bulbs consume 0.5 W. How many hours will the flashlight last?

Answers

Answer:

time = 5.22 hr

Explanation:

Given data:

Energy of battery = 9400 J

Power consumed by three led bulb is 0.5 watt

we know Power is give as

[tex]Power = \frac{ energy}{time}[/tex]

plugging all value and solve for time

[tex]time = \frac{energy}{power}[/tex]

[tex]time = \frac{9400}{0.5}[/tex]

time = 18,800 sec

in hour

1 hour = 3600 sec

therefore in 18,800 sec

[tex]time = \frac{18800}{3600} = 5.22 hr[/tex]

a 100mh inductor is placed parallel with a 100 ohm resistor in the circuit , the circuit has a source voltage of 30 vac and a frequancy of 200 Hz what is the current through the inductor

Answers

Answer:

current through the inductor  = 0.24 A

Explanation:

given data

inductor = 100mh

resistor = 100 ohm

voltage = 30 vac

frequancy = 200 Hz

to find out

current through the inductor

solution

current through the inductor will be when inductor is place parallel with resistor

across resistor

XL = i2πl

XL = i2π×200×100×[tex]10^{-3}[/tex] = i126.66

so

current through the inductor = [tex]\frac{voltage}{XL}[/tex]

current through the inductor = [tex]\frac{30}{125.66}[/tex]

current through the inductor  = 0.24 A

Write Python expressions using s1, s2, and s3 and operators and * that evaluate to: (a) 'ant bat cod'

Answers

Final answer:

To form the string 'ant bat cod', you can concatenate the strings s1, s2, and s3 using the + operator or repeat each string based on the desired repetition using the * operator.

Explanation:

Python expressions using s1, s2, and s3 and operators and * that evaluate to: (a) 'ant bat cod'.

s1 + ' ' + s2 + ' ' + s3s1 * 1 + ' ' + s2 * 2 + ' ' + s3 * 3

Determine the percent increase in the nominal moment capacity of the section in Problem 2 when including compression steel at top equal to 0.5 the area of the tension steel at the bottom.

Answers

Explanation:

Please kindly share your problem two with us as to know the actual problem we are dealing with, the question looks incomplete

A glass window pane, 1 m wide, 1.5 m high, and 5 mm thick, has a thermal conductivity of kg = 1.4 W/(m∙K). On a cold winter day, the indoor and outdoor temperatures are 15 °C and −25 °C respectively. (a) For a single-pane window at steady state, what is the rate of heat transfer through the glass? (10 pts) (b) To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining panes are separated by a dead-air space. The thermal conductivity of air is ka = 0.024 W/(m∙K). If the spacing between the two glasses is 10 mm. Calculate the temperatures of the glass surfaces in contact with the dead-air at the steady state. (10 pts) (c) Calculate the heat loss through the double-pane window in (b). (5 pts)

Answers

Answer:

a) Rate of heat transfer = 16.8KW

b) Temperature of glass surface = 15degree Celsius

c) Heat loss through frame = 141.34W

Explanation:

The concept used to approach this question is the Fourier's law of head conduction postulated by Joseph Fourier. it states that the rate of heat flow through a single homogeneous solid is directly proportional to the area and to the direction o heat flow and to the change in temperature with respect to the path length. Mathematically,

Q = -KA dt/dx

The detailed and step by step calculation is attached below

A) Fix any errors to get the following program to run in your environment. B) Document each line of code with comments and describe any changes you had to make to the original code to get it to work. C) Write a summary of what your final version of the program does.

You may also add white space or reorder the code to suit your own style as long as the intended function does not change.

#include
using namespace std;

void m(int, int []);
void p(const int list[], int arraySize)
{
list[0] = 100;
}

int main()
{
int x = 1;
int y[10];
y[0] = 1;

m(x, y);

cout << "x is " << x << endl;
cout << "y[0] is " << y[0] << endl;

return 0;
}

void m(int number, int numbers[])
{
number = 1001;
numbers[0] = 5555;
}

Answers

Answer:

Answer is explained below

Explanation:

Part A -:

Error statement -:

/*

prog.cpp: In function ‘void p(const int*, int)’:

prog.cpp:7:15: error: assignment of read-only location ‘* list’

list[0] = 100;

*/

There is one error in the code in following part

void p( const int list[], int arraySize)

{

list[0] = 100;

}

you are passing list as constant but changing it inside the function that is not allowed. if you pass any argument as const then you can't change it inside the function. so remove const from function argument solve the error.

Part B -:

change made

void p( int list[], int arraySize)

{

list[0] = 100;

}

Executable fully commented code -:

#include <iostream> // importing the iostream library

using namespace std;

void m(int, int []); // Function declearation of m

void p( int list[], int arraySize) // definition of Function p

{

list[0] = 100; // making value of first element of list as 100

}

int main()

{

int x = 1; // initilizing x with 1

int y[10]; // y is a array of 10 elements

y[0] = 1; // first element of y array is 1

m(x, y); // call m function

// printing the desired result

cout << "x is " << x << endl;

cout << "y[0] is " << y[0] << endl;

return 0;

}

void m(int number, int numbers[]) // Function definition of m

{

number = 1001; // value of number is 1001 locally

numbers[0] = 5555; // making value of first element of numbers array 5555

}

Part C :-

In program we initilize x with value 1 and create an array y of 10 elements.

we initilize the y[0] with 1\

then we call function m. In function m ,first argument is value of x and second argument is the pointer to the first element of array y.

so value of x is changed locally in function m and change is not reflected in main function.

but value of y[0] is changed to 5555 because of pass by refrence.

So we are getting the following result :-

x is 1

y[0] is 5555

It is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water entering at 20 °C?

Answers

Answer:

Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.

Consider a point in a structural member that is subjected to plane stress. Normal and shear stress magnitudes acting on horizontal and vertical planes at the point are Sx = 175 MPa, Sy = 90 MPa, and Sxy = 75 MPa.

Answers

Answer:

C = 132.5 MPa

R = 86.20 MPa

Explanation:

Given

σx = 175 MPa

σy = 90 MPa

τxy = 75 MPa

For the given state of stress at a point in a structural member, determine the center C and the radius R of Mohr’s circle.

We apply the following equation for the center C

C = (σx + σy) / 2

C = (175 MPa + 90 MPa) / 2

C = 132.5 MPa

The Radius can be obtained as follows

R = √(((σx - σy) / 2)² + (τxy)²)

R = √(((175 MPa - 90 MPa) / 2)² + (75 MPa)²)

R = 86.20 MPa

If a barrel of oil weighs 1.5 kN, calculate the specific weight, density, and specific gravity of the oil. The barrel weighs 110 N

Answers

Answer

given,

oil barrel weight  = 1.5 k N = 1500 N

weight of the barrel = 110 N

Assuming volume of barrel = 0.159 m³

weight of oil = 1500-110

                     = 1390 N

[tex]specific\ weight = \dfrac{weight}{volume}[/tex]

[tex]specific\ weight = \dfrac{1390}{0.159}[/tex]

            = 8742.14 N/m³

[tex]mass = \dfrac{weight}{g}[/tex]

[tex]mass = \dfrac{1390}{9.8}[/tex]

              = 141.84 kg

[tex]density = \dfrac{mass}{volume}[/tex]

[tex]density = \dfrac{141.84}{0.159}[/tex]

                    = 892.05 kg/m³

[tex]Specific\ gravity = \dfrac{density\ of\ oil}{density\ of\ water}[/tex]

[tex]Specific\ gravity = \dfrac{892.05}{1000}[/tex]

                    = 0.892

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?

Answers

Final answer:

The answer explains the energy of a quantum harmonic oscillator, calculates the frequency of an emitted photon, and identifies the region of the electromagnetic spectrum it belongs to.

Explanation:

a. Energy: The energy of a quantum harmonic oscillator can be represented as En = (n+1/2)h(sqrt(k/M)), where n = 0,1,2... and h represents Planck's constant.

b. Frequency Calculation: Using the given values of k = 15 N/m and m = 4 x 10^-26 kg, you can calculate the frequency of the emitted photon using the formula w = sqrt(k/M)/(2pi).

c. Electromagnetic Spectrum: To determine the region of the electromagnetic spectrum the photon belongs to, compare the frequency calculated to the known ranges of various regions like x-ray, visible, and microwave.

Electrical failure can lead to ________failure, which in turn can lead to software failure.

Answers

Answer:

Electrical failure can lead to _hardware__failure, which in turn can lead to software failure

Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 300°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 3560 kJ and kinetic and potential energy effects are negligible.
Determine the final volume, in m3, and the work for the process, in kJ.

Answers

Final answer:

To determine the final volume and work done during the heating of water in a piston-cylinder assembly at constant pressure, tabulated data like steam tables are required since water is not an ideal gas. The work done is calculated using the formula W = PΔV.

Explanation:

The student has been asked to find the final volume and the work done during a constant pressure process in which 5 kg of water is heated in a piston-cylinder assembly from an initial state of 5 bar and 300°C. To solve for the final volume and work done, one would typically use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat input minus the work output. However, since water at this state is not an ideal gas, tabulated data from steam tables or software would be used to determine the specific volume at the final state and then multiplied by the mass to find the total volume. The work done in a constant pressure process is equal to the pressure times the change in volume (W = PΔV). Without the final specific volume from the tables, we cannot compute the final volume or work directly.

Consider a steady flow Carnot cycle with water as the working fluid. The maximum and minimum temperatures in the cycle are 350 and 50 C. The quality of water is 0.891 at the beginning of the heat rejection process and 0.1 at the end. Determine:

(a) the thermal efficiency (how much percent).
(b) the pressure at the turbine inlet, and
(c) the network output

a) Etath,C= %
b) P2=MPa
c) wnet = kJ/kg

Answers

Answer:

a) Etath = 48.2 %

b) P2 = 1.195 MPa

c) wnet = 1749.14 KJ/kg

Explanation:

Given:

T1 = T2 = 50 , TL = 50 + 273 = 323 K

T3 = T4 = 350 , TH = 350 + 273 = 623 K

x3 =0.891

x4 = 0.1

Part a

Thermal efficiency (Etath) of carnot cycle is:

Etath = 1 - (TL / TH )

Etath = 1 - (323) / (623) = 48.2 %

Part b

Note:

From A - 4 Table

T1 = 50 C @sat

sf = 0.7038 KJ/kg.K

sfg = 7.3710 KJ/Kg.K

s2 = s3 = sf + x3 * (sfg)

s2 = s3 = 0.7038 + 0.891*7.3710 = 7.271361 KJ/kg.K

Thus,

@ T2 = 350 C  ; sg = 5.2114 KJ/kg.K

s2 = 7.271361 KJ/kg.K

s2 > sg@T = 350 C, hence super-heated region.

Use Table A-6

T2 = 350 C

s2 = 7.271361 KJ/kg.K

P2 = 1.195 MPa (using interpolation)

part c

wnet can be calculated by finding the enclosed area on a T-s diagram:

s4= sf + x4*sfg =  0.7038 + 0.1*7.3710 = 1.4409 KJ/kg.K

wnet = (TH - TL)*(s3 - s4)

wnet = (623 - 223) * (7.271361 - 1.4409) = 1749.14 KJ/kg

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