Calculate the change in enthalpy of 1.94 mol of PbO(s) if it is cooled from 732 K to 234 K at constant pressure.

Answer:

The case with replacement has higher entropy

Explanation:

The complete question is given:

'Drawing with and without replacement. An urn contains r red,  w white  and b black balls. Which has higher entropy, drawing k ≥ 2 balls from the urn with  replacement or without replacement?'

Solution:

- n drawing is the same irrespective of whether there is replacement or not.

-  X to denotes drawing from an urn with r red balls,  w white balls and b black balls. So, n = b + r +  w.

We have:

                                      p_X(cr) = r / n

                                      p_X(cw) = w / n

                                      p_X(cb) = b / n

- Now, if  Xi is the ith drawing with replacement then Xi are independent and p_Xi (x) = pX(x) for x e ( cr , cb , cw ).

- Now, let  Yi be the ith drawing with replacement where Yi are not independent p_Yi (x) = pX(x) for x ∈ X.

- To see this, note  Y1 =  X and assume it is true for  Yi and consider  Yi+1:

    p_Y(i+1) (cr) = p(Y(i+1),Yi).(cr, cr) + p(Y(i+1),Yi).(cr, cw) + p(Y(i+1),Yi).(cr, cb)

= pY(i+1)|Yi  (cr|cr)*pYi  (cr) +pY(i+1)|Yi  (cr|cw)*pYi (cw) + pY(i+1)|Yi  (cr|cb)*pYi (cb)

= r*( r - 1 )/n*(n-1) + w*r/n*(n-1) + b*r/n*(n-1) = r / n =  p_X(cr)

- This means, using the chain rule and the conditioning theore m:

H(Y1, Y2, . . . , Yn) =  H(Y1) +  H(Y2|Y1) +  H(Y3|Y2, Y1) + ... H(Yn|Yn−1, . . . , Y1)

=< SUM H(Yi) = n*H(X) =  H(X1, X2, . . . , Xn)

- with equality if and only if the  Yi were independent:

                          H(Y1, Y2, . . . , Yn) < H(X1, X2, . . . , Xn)

Answer: The case with replacement has higher entropy

Answer:

A) while (num >= 0)

Explanation:

To understand why we need to focus on the module and division operation inside the loop. num % 10 divide the number by ten and take its remainder to then add this remainder to sum, the important here is that we are adding up the number in reverse order and wee need to repeat this process until we get the first number (1%10 = 1), therefore, num need to be one to compute the last operation.

A) this is the correct option because num = 1 > 0 and the last operation will be performed, and after the last operation, num = 1 will be divided by 10 resulting in 0 and 0 is not greater than 0, therefore, the cycle end and the result will be printed.

B) This can not be the option because this way the program will never ends -> 0%10 = 0 and num = 0/10 = 0

C) This can not be the option because num = 1 > 1 will produce an early end of the loop printing an incomplete result

D) The same problem than C

E) There is a point, before the operations finish, where sum > num, this will produce an early end of the loop, printing an incomplete result

Loops are program statements that are used to carry out repetitive and iterative operations

The missing loop header is (a) while (num > 0)

To calculate the sum of the digits of variable num, the following must be set to be true

The loop header must be set to keep repeating the loop operations as long as the value of variable num is more than 0

To achieve this, we make use of while loop,

And the loop condition (as described above) would be num > 0

Hence, the true option is (a) while (num > 0)

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In the process of analyzing a thermodynamic system it is important to identify what system is being worked on and the processes and properties if the system

The magnitude of the force acting on the shaft, is approximately 1,336.5 N

The reason the value for the force magnitude acting on the shaft is correct is as follows:

The known parameters are:

The cross-sectional area of the shaft, Aₐ = 0.8 cm²

The required gas pressure in the cylinder, P = 3 bar

The mass of the piston, m₁ = 24.5 kg

The mass of the shaft, m₂ = 0.5 kg

The diameter of the piston, D = 10 cm

The atmospheric pressure, Pₐ = 1 bar

Required:

The magnitude of the force F acting on the shaft

Solution:

The force due to the gas in the cylinder, [tex]\mathbf{F_{gas}}[/tex], is given as follows;

[tex]F_{gas}[/tex] = 3 bar × π × (10 cm)²/4 = 2,359.19449 N

The force due to the atmosphere, [tex]\mathbf{F_{atm}}[/tex], is given as follows;

[tex]F_{atm}[/tex] = 1 bar × ((π × (10 cm)²/4) -  0.8 cm²) ≈ 777.4 N

The force due to the piston and shaft, [tex]\mathbf{F_{ps}}[/tex], is given as follows;

[tex]F_{ps}[/tex] = (24.5 kg + 0.5 kg) × 9.81 m/s² = 245.25 N

The magnitude of the force acting on the shaft, F = [tex]F_{gas}[/tex] - ([tex]\mathbf{F_{atm}}[/tex] + [tex]\mathbf{F_{ps}}[/tex])

∴ F = 2,359.19449 N - (777.4 N + 245.25 N) ≈ 1,336.5449 N

The magnitude of the force acting on the shaft, F ≈ 1,336.5 N

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Answer:

SaaS

Explanation:

Software as a service (SaaS) is also called software on demand, it involves a third party that centrally hosts the software and provides it to the end user.

All aspects of hosting is handled by the third party: application, data, runtime, middleware, operating system, server, virtualization, storage and networking are all handled by the provider.

This is an ideal software service for Fictional corp, as there will be no need to hire additional IT staff to maintain the new CRM software.

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.

The values of the potential differences for the three questions are;

A) [tex]V_{MN} = -147 V[/tex]

B) [tex]V_{MQ}[/tex] = -120 V

C) [tex]V_{N} = 51 V[/tex]

We are given the expression of the electric field as;

E = (6x² x^ + 6y y^ +4 z^) V/m

[tex]V_{MN} = -\int\limits^M_N {E} \, dx[/tex]

Integrating this with the M and N coordinates as boundaries in mind gives;

[tex]V_{MN} = -[6\frac{x^{3}}{3} + 6\frac{y^{2}}{2} + 4z]^{2,6,1}_{-3,-3,2}[/tex]

[tex]V_{MN} = -[2{x^{3} + 3y^{2} + 4z]^{2,6,1}_{-3,-3,2}[/tex]

Plugging in those boundary values and solving using an online integral calculator gives;

[tex]V_{MN} = -147 V[/tex]

[tex]V_{MQ} = -[6\frac{x^{3}}{3} + 6\frac{y^{2}}{2} + 4z]^{2,6,1}_{4,-2,-35}[/tex]

[tex]V_{MQ} = -[2{x^{3} + 3y^{2} + 4z]^{2,6,1}_{4,-2,-35}[/tex]

Plugging in those boundary values and solving using an online integral calculator gives;

[tex]V_{MQ}[/tex] = -120 V

C) We are told that V = 2 at P(1,2,4). Thus potential difference between point V and N is;

[tex]V_{NP} = -[6\frac{x^{3}}{3} + 6\frac{y^{2}}{2} + 4z]^{-3,-3,2}_{1,2,4}[/tex]

[tex]V_{NP} = -[2{x^{3} + 3y^{2} + 4z]^{-3,-3,2}_{1,2,4}[/tex]

Plugging in those boundary values and solving using an online integral calculator gives; [tex]V_{NP} = 49 V[/tex]

Thus;

[tex]V_{N} = V + V_{NP}[/tex]

[tex]V_{N}[/tex] = 2 + 49

[tex]V_{N} = 51 V[/tex]

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Answer:

Energy Transfer  =  350 kJ

Explanation:

The net work can be determined from an energy balance. That is, with assumption

∆KE + ∆PE + ∆U = Q − W

Where

∆KE = ∆PE = 0 (Since There is no significant change in the kinetic or potential energy of the steam)

The net work is the sum of the work associated with the paddlewheel Wpw

and the work done on the piston Wpiston:

W = Wpw + Wpiston

From the given information, Wpw= −18.5 kJ,

Collecting results:

Wpw + Wpiston = Q − ∆U

Wpiston = Q − ∆U − Wpw= Q − m (u2− u1) − Wpw

Where Q=80kJ, m=5kg, u2 = 2659.6 kJ/kg,  u1 = 2709.9 kJ/kg

= 80 kJ − 5 kg (2659.6 − 2709.9)kJ/kg − ( −18. 5 kJ)

= 350 kJ

The energy transfer by work from steam to piston is ; [tex]W_{p}[/tex] =  350 kJ

Given that ;

ΔU = ( 2659.6 - 2709.9 ) = - 50.3 kJ/Kg

Q ( heat magnitude ) = 80 kJ

m ( mass of steam ) = 5 kg

Energy transferred by paddle wheel ( [tex]W_{pp}[/tex] ) = - 18.5 kJ

Energy transferred to piston ( [tex]W_{p}[/tex] ) = ?

Total work done given that there is no change on K.E. and P.E.

Total work done = m ( ∆U ) = Q - W  ----- ( 1 )

where W = [tex]W_{pp} + W_{p}[/tex]

Equation ( 1 ) becomes

ΔU = Q - [tex]( W_{PP} + W_{P} )[/tex] ------ ( 2 )

Therefore the energy transferred to piston by work from steam ( [tex]W_{p}[/tex] )

[tex]W_{p}[/tex] = Q - m( ΔU )  - [tex]W_{pp}[/tex]

     = 80 - 5(- 50.3 ) - ( -18.5 )

     = 350 kJ

Hence the The energy transfer by work from steam to piston is ; [tex]W_{p}[/tex] =  350 kJ

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Answer:

The maximum temperature of the cycle is 1065⁰R

Explanation:

The maximum temperature in degree Rankin can be obtained using the formula below;

[tex]\frac{T_2}{T_1} =[\frac{V_1}{V_2}]^{1.4-1}[/tex]

Where;

T₂ is the  maximum temperature of the cycle

T₁ is the initial temperature of the cycle = 520 deg R = 520 ⁰R

V₁/V₂ is the compression ratio = 6

[tex]T_2 = T_1(\frac{V_1}{V_2})^{0.4}[/tex]

[tex]T_2=T_1(6)^{0.4}[/tex]

[tex]T_2=520^0R(6)^{0.4}[/tex]

T₂ = 1064.96 ⁰R

Therefore, the maximum temperature of the cycle is 1065⁰R

Answer:

Explanation:

# taking first number

low = int(input("Enter a number: "))

# if that is valid

if low >= 0:

# considering it as high, sum and input

high = low

sum = low

count = 1

inp = low

# breaking if negative number is entered

while True:

# taking user input of numbers

inp = int(input("Enter a number: "))

if inp >= 0:

# adding it to sum

sum = sum + inp

# checking for low

if low > inp:

low = inp

# checking for high

if high < inp:

high = inp

# tracking count

count = count + 1

else:

break

# printing output

print("\nSum =",sum)

print("count =",count)

print("Average =",round(sum/float(count),2))

print("Lowest =",low)

print("Highest =",high)

# no valid numbers

else:

print("no valid numbers entered")

Answer:

Explanation:

A- Specific gravity and Absorption Test: Specific gravity is a measure of a material’s density as compared to the density of water at 73.4°F (23°C). Therefore, by definition, water at a temperature of 73.4°F (23°C) has a specific gravity of 1. Absorption is also determined by the same test procedure and it is a measure of the amount of water that an aggregate can absorb into its pore structure.

Specific gravity is used in a number of applications including Superpave mix design, deleterious particle identification and separation and material property change identification while

B- Soundness Test : This determines an aggregate's resistance to disintegration by weathering and in particular, freeze-thaw cycles. Aggregates that are durable (resistant to weathering) are less likely to degrade in the field and cause premature HMA pavement distress and potentially failure.It is used to identify the excess amount of lime in cement.

C - Sieve analysis Test: is a practice or procedure used to assess the particle size distribution (also called gradation) of a granular material by allowing the material to pass through a series of sieves of progressively smaller mesh size and weighing the amount of material that is stopped by each sieve as a fraction of the whole mass. This test is used to describe the properties of the aggregate and to see if it is appropriate for various civil engineering purposes such as selecting the appropriate aggregate for concrete mixes and asphalt mixes as well as sizing of water production well screens.

Answer:

Detailed Answer is given below,

Explanation:

(a) Elastic strain

Elastic deformation is defined as the limit for the deformation values up to which the object will bounce and return to the original shape when the load is removed.

When the object is subjected to an external load, it deforms. If the load exceeds the load corresponding to the elastic limit of the object, then, after removing the load, the object cannot return to its original geometric specification.

(b) Plastic strain

Plastic Strain is a deformation that cannot be recovered after eliminating the deforming force. In other words, if the applied tension is greater than the elastic limit of the material, there will be a permanent deformation.

(c) Creep strain

The deformation of the material at a higher temperature is much more than at the normal temperature known as creep.

Subsequently, the deformation produced in heated material is called creep deformation. Creep deformation is a function of temperature.

(d) Tensile viscosity

Viscosity is a main parameter when measuring flux fluids, such as liquids, semi-solids, gases and even solids. Brookfield deals with liquids and semi-solids. Viscosity measurements are made together with the quality and efficiency of the product. Any person involved in the characterization of the flow, in research or development, quality control or transfer of fluids, at one time or another is involved with some kind of viscosity measurement.

(e) Recovery

Recovery is a process whereby deformed grains can reduce their stored energy by eliminating or reorganizing defects in their crystalline structure. These defects, mainly dislocations, are introduced by plastic deformation of the material and act to increase the elastic limit of a material.

(f) Relaxation

Stress relaxation occurs in polymers when they remain tense for long periods of time.

These alloys have very good resistance to stress relaxation and, therefore, are used as spring materials.

Stress relaxation is a gradual reduction of stress over time in constant tension.

Answer:

i. Utility

ii. Performance

Explanation:

While there may be other vulnerabilities of the iostream library when compared to other C++ libraries, the two most common vulnerabilities are

I. Utility

II. Performance

Utility

The capacity of iostream to extend its structured read and write functions is its biggest features. One can overload the operator "<<" for various functions and types and simply use them.

This can't be done with fprintf but it can be used for classes in namespaces. Also, new streambuf types and even streams can't be created just anytime.

Performance

The effect of, “iostreams is intended to do far more than C-standard file IO.” but that is not always true because with iostreams, though there is an extensible mechanism for writing any type directly to a stream, one can't easily write new streambuf’s that will allow you to (via runtime polymorphism) be able to work with existing code.

Answer:

Power of the motor that kept the wheel rotating at 12 rev/hr despite friction is 342.79W.

Explanation:

Pls refer to the attached file. The explanation is long to pen down here.

Answer:

a.The current is cut in half

b.the current stays the same

c.the current doubles

d.the current becomes zero

Explanation:

The current doubles

Answer:

The main purpose the various types of drawing used for the design and erection of steel framed buildings is to successfully construct a solid and strong building

Explanation:

The main purpose the various types of drawing used for the design and erection of steel framed buildings is to successfully construct a solid and strong buildings.

Steel forms the skeleton of a building, essentially the part of the building that holds everything up and together. Steel has so many advantages when compared to other structural building material such as concrete, plastic, timber and composite materials.

Answer:

T = 3600 lb

Explanation:

Given:

- coefficient of static friction @a u_a = 0.2

- coefficient of static friction @b u_b = 0.3

- Weight of the loaded bin W = 8500 lb

Find:

- Find the force in the cable needed to begin the lift.

Solution:

- Draw the forces on the diagram. see attachment.

- Take sum of moments about point B as zero:

                     (M)_b =   W*12 - N_a * 22 = 0

                      N_a = W*12 / 22 = 8500*12 / 22

                      N_a = 4636.364 lb

- Compute friction force F_a @ point A:

                      F_a = u_a*N_a = 4636.364*0.2

                      F_a = 927.2727 lb

- Take sum of moments about point A as zero:

                     -W*10 - F_b*sin(30)*22+ 22*N_b*cos(30) + 22*T*sin(30) = 0

Where,           F_b = u_b*N_b = N_b*0.3            

Hence,           -85000 - 3.3*N_b + 11sqrt(3)*N_b + 11 T = 0

                      15.753*N_b + 11*T = 85000    ...... 1    

- Take sum of forces in x-direction equal to zero:

                      T*cos(30) - N_b*sin(30) - u_b*N_b*cos(30) - F_a = 0

                      T*cos(30) - 0.75981*N_b = 927.2727   ..... 2

- Solve two equation simultaneously:

                      T = 3600 lb , N_b = 2882 lb

Answer:

not provide the sprinklers, then the type of construction will be Type II B under IBC

Explanation:

given data

3 story office building = 20,000 square feet per floor

solution

we know when sprinkler is provide in high rise building to resist fire

and it provide in building as floor area exceed allowable permissible area of  building as IBC

so IBC for Type II B allowable area =  19000 square feet per floor

and type III B allowable area =  23000 square feet per floor

so when we design the building by type III B construction, the sprinklers require to provide

but not provide the sprinklers, then the type of construction will be Type II B under IBC

Answer:

The PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

Explanation:

From table 3.5 of Traffic Engineering by Mannering

R_v=835

R=835+(10ft/2)= 840 ft.

Now T is given as

T=R tan(Δ/2)

Here Δ is the central angle of curve given as 35°

So

T=R tan(Δ/2)

T=840 x tan(35/2)

T=840 x tan(17.5)

T=264.85

Now

STA PC=482+72-(2+64.85)=480+07.15

Also L is given as

L=(π/180)RΔ

Here R is the radius calculated as 840 ft, Δ is the angle given as 35°.

L=(π/180)RΔ

L=(π/180)x840 x35

L=512.87 ft

STA PT=480+07.15+5+12.87=485+20.02

Now Ms is the minimum distance which is given as

[tex]M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\[/tex]

Here R_v is given as 835

SSD for 50 mi/hr is given as 425 ft from table 3.1 of Traffic Engineering by Mannering

So Ms is

[tex]M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\M_s=835(1-cos(\frac{90 \times 425}{\pi 835}))\\M_s=26.92 ft[/tex]

Now for the clearance from the inside lane

Ms=Ms-lane length

Ms=26.92-5= 21.92 ft.

So the PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

Answer:

The motor inertia is 7.958 X 10⁻² kg.m²

Explanation:

To determine the motor inertia, the following formula applies.

Neglecting the damping effect,

[tex]T = \frac{J}{9.55}.\frac{\delta n}{\delta T}[/tex]

Where;

T is the constant torque applied to the motor = 5Nm

J is the motor inertia = ?

δn is the change in angular speed of the motor = 1800 r/min

δT is change in time of the unloaded motor from rest = 3 sec

[tex]J = \frac{9.55* \delta T* T}{\delta n}[/tex]

[tex]J = \frac{9.55* 3* 5}{1800}[/tex] = 0.07958 kg.m² = 7.958 X 10⁻² kg.m²

Therefore, the motor inertia is 7.958 X 10⁻² kg.m²

Answer:

Option A

Explanation:

Alloys are metal compounds with two or more metals or non metals to create new compounds that exhibit superior structural properties. Alloys have high level of hardness that resists deformation thereby making it less ductile compared to polymers. This is due to the varying difference in the chemical and physical characteristics of the constituent metals in the alloy.

Answer:

“height is a quantitative variable ”

Explanation:

According to the question asked, answer is “height is a quantitative variable ”

Height is a quantitative variable because it is related to the measurement and in measurement, when we measure something we deal with number (numerical data)

Numerical data is a type of quantitative data that is why we say “height is a quantitative variable”  

There are some other possible questions in the given paragraph which I would like to mention here,  are as following:

Which are the categorical variables in the given report?

Answer: Energy star complaints

Top, Bottom or side-by-side freezer

Which are the quantitative variables in the given report?

Answer: Estimated Energy Consumption in kilowatts

Width, depth, and height in inches

Capacity in Cubic Feet  

What are the individuals in the report?

Answer: The brand name and model  

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = [tex]\frac{NE}{t}[/tex]

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = [tex]\frac{NE}{t}[/tex] = [tex]\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}[/tex]          [t = 1 s for per second]

or

550 × 10³ = [tex]\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}[/tex]

or

550 = N × 4903.24 × 10⁻³⁴

or

N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second

The number of photons that are emitted by this AM radio transmitter is equal to [tex]1.12 \times 10^{33}\;photons.[/tex]

Given the following data:

Scientific data:

In order to determine the number of photons that are being emitted by this AM radio transmitter, we would solve for the quantity of energy it consumes by using Planck-Einstein's equation.

Mathematically, the Planck-Einstein relation is given by the formula:

[tex]E = hf[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]E = 6.626 \times 10^{-34}\times 740 \times 10^3\\\\E = 4.903 \times 10^{-28}\;Joules.[/tex]

For the number of photons:

[tex]n=\frac{Power}{Energy} \\\\n=\frac{550 \times 10^3}{4.903 \times 10^{-28}} \\\\n=1.12 \times 10^{33}\;photons.[/tex]

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The total work required to pump all the water out of the swimming pool over the top is 3,744,000 foot-pounds.

Define the Variables

The density of water,[tex]\( \delta \)[/tex], is 62.4 lb/ft³.

The pool's dimensions are 50 ft long (x-direction), 25 ft wide (y-direction), and filled to 8 ft deep (z-direction).

Setup the Integral

Volume of a slice of water at depth*:  

The slice at depth zis a horizontal slice of water with thickness dz and area:
=50 x 25

= 1,250 ft².

Weight of the slice of water:  

[tex]\[ dW = \delta \times \text{Volume} = 62.4 \times 1250 \times dz \text{ lb} \][/tex]

Height the water needs to be lifted:  

The water at depth z needs to be lifted to the rim of the pool, which is 10 ft above the bottom. Thus, each slice is lifted 10 - zfeet.

Work to lift this slice of water:  

[tex]\[ dU = dW \times \text{Height} = 62.4 \times 1250 \times (10 - z) \times dz \text{ ft-lb} \][/tex]

Integrate

To find the total work, integrate [tex]\( dU \)[/tex] from z = 0 to z = 8 ft (since the water depth is 8 ft):

[tex]\[ U = \int_0^8 62.4 \times 1250 \times (10 - z) \, dz \text{ ft-lb} \][/tex]

Calculate the Integral

[tex]\[ U = 62.4 \times 1250 \int_0^8 (10 - z) \, dz \][/tex]

Compute the integral:

[tex]\[ \int_0^8 (10 - z) \, dz = [10z - \frac{1}{2}z^2]_0^8 \\= [80 - \frac{1}{2}(64)] \\= 80 - 32 \\= 48 \][/tex]

[tex]\[ U = 62.4 \times 1250 \times 48 \\= 3,744,000 \text{ ft-lb} \][/tex]

Answer:

A. 36 minutes

B. 120 minutes

C.

i. 144 minutes

ii. 984 minutes

D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.

E. 98.33km

Explanation

A.

Given

dAb = 10km

dBc = 10km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 0.2 hours

= 24 minutes + 0.2 * 60 minutes

= 24 minutes + 12 minutes

= 36 minutes

B.

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 0.2 hours

= 3 minutes + 0.2 * 60 minutes

= 3 minutes + 12 minutes

= 15 minutes

Total End delay for 8 cars = 8 * 15 = 120 minutes

C.

Given

dAb = 100km

dBc = 100km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

i. Cars travelling together

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 2 hours

= 24 minutes + 2 * 60 minutes

= 24 minutes + 120 minutes

= 144 minutes

ii. Cars travelling separately

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 2 hours

= 3 minutes + 2 * 60 minutes

= 3 minutes + 120 minutes

= 123 minutes

Total End delay for 8 cars = 8 * 123 = 984 minutes

D.

Distance = 100km

Time = 1 min/car

Car 1 is 1 minute ahead of car 2 --- at toll A and B

If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes

Time delay = 11 - 10 = 1 minute

Distance = time * speed

= 1 minute * 100km/hr

= 1 hr/60 * 100 km/hr

= 100/60

= 1.67km

E.

Given

Distance = 100km

Distance behind = 1.67

Maximum value of dBc = 100km - 1.67km = 98.33km

The maximum distance that can be reached is 98.33km

Answer:

net amount of energy change of the air in the room during a 30-min period = 660KJ

Explanation:

The detailed calculation is as shown in the attached file.

Answer:

660KJ

Explanation:

Given

Let Q = Heat Loss from room = 50kj/min

Let W = Work Supplied to room = 1.2KW

1 kilowatt = 1 kilojoules per second

So, W = 1.2KJ/s

In heat and work (Sign Convention)

We know that

1. Heat takes positive sign when it is added to the system

2. Heat takes negative sign when it is removed from the system.

3. Work done is considered positive when work is done by the system

4. Work done is considered negative when work is done on the system.

From the above illustration, heat loss (Q) = -50KJ/Min

In 30 minutes time, Q = -50Kj/Min * 30 Min

Work done in 30 minutes = -1500 KJ

Also, work supplied = -1.2Kj/s

Work supplied to the system in 30 minutes = -1.2Kj/s * 30 minutes

W = -1.2 KJ/s * 30 * 60 seconds

W = -2160KJ

In thermodynamics (First Law)

Q = W + ΔU

-1500 = -2160 + ΔU

∆U = 2160 - 1500

∆U = 660KJ

The spacing between two adjacent bright fringes, when a helium-neon laser with a wavelength of 633 nm illuminates two slits 0.50 mm apart and a screen is placed 2.5 m behind the slits, is 3.16 mm.

To answer the question on the spacing between two adjacent bright fringes for a helium-neon laser with a wavelength (λ) of 633 nm illuminating two slits spaced 0.50 mm apart, with a viewing screen 2.5 m behind the slits, we use the formula for the fringe spacing in a double-slit interference pattern, Δy = (λL) / d, where Δy is the fringe spacing, λ is the wavelength of the light, L is the distance from the slits to the screen, and d is the distance between the slits.

Given λ = 633 nm = 633 × 10-9 m, L = 2.5 m, and d = 0.50 mm = 0.50 × 10-3 m, plugging these values into the formula gives:

Δy = (633 × 10-9 m × 2.5 m) / (0.50 × 10-3 m) = 3.16 × 10-3 m = 3.16 mm

Therefore, the spacing between two adjacent bright fringes is 3.16 mm.

Answer:

Explanation:

val db = ("John", "x3456", 50.1) :: ("Jane", "x1234", 107.3) ::

        ("Joan", "unlisted", 12.7) :: Nil

type listOfTuples = List[(String, String, Double)]

def find_salary(name: String) = {

 def search(t: listOfTuples): Double = t match {

   case (name_, _, salary) :: t if name == name_ => salary

   case _ :: t => search(t)

   case Nil    =>

     throw new Exception("Invalid Argument in find_salary")

 }

 search(db)

}

def select(pred: (String, String, Double) => Boolean) = {

 def search(found: listOfTuples): listOfTuples = found match {

   case (p1, p2, p3) :: t if pred(p1, p2, p3)  => (p1, p2, p3) :: search(t)

   case (p1, p2, p3) :: t if !pred(p1, p2, p3) => search(t)

   case Nil => Nil

   case _ => throw new Exception("Invalid Argument in select function")

 }

 search(db)

}

println("Searching the salary of 'Joan' at db: " + find_salary("Joan"))

println("")

val predicate = (_:String, _:String, salary:Double) => (salary < 100.0)

println("All employees that match with predicate 'salary < 100.0': ")

println("\t" + select(predicate) + "\n")

Answer:

b, c, and e

Explanation:

The values that are used for case labels must be a constant expression.

Let's examine the options;

a. five -> declared as just int

b. FIVE -> declared as constant expression

c. 5 -> It is a constant expression

d. five++ -> incremented version of option a

e. FIVE+1 -> incremented version of option b

Answer:

box 2=highest

box3= 2

box 1=3

box 4=lowest

Explanation:

Decide how the sketches below would be listed, if they were listed in order of decreasing force between the charges. That is, select "1" beside the sketch with the strongest force between the charges, select "2" beside the sketch with the next strongest force between the charges, and so on.

take note that like charges repel while unlike charges attract

from the law of electric attraction, we know that

f=kQq/r^2

force is directly proportional to the charges,

-3-1= has the highest force of repulsion

the first bos, the balls are -2-1=-3 they are repulsive

second box=-3-1=-4

third box=-3-1=-4

fourth=-1-1=-2

box 2=highest

box3= 2

box 1=3

box 4=lowest

Answer:

The number of kilo- grams of hydrogen that pass per hour through this sheet of palladium is [tex]4.1 * 10^{-3} \frac{kg}{h}[/tex]

Explanation:

Given

x1 = 0 mm

x2 = 6 mm = 6 * [tex]10^{-3}[/tex] m

c1 = 2 kg/[tex]m^{3}[/tex]

c2 = 0.4 kg/[tex]m^{3}[/tex]

T = 600 °C

Area = 0.25 [tex]m^{2}[/tex]

D = 1.7 * [tex]10^{8} m^{2}/s[/tex]

First equation

J = - D [tex]\frac{c1 - c2}{x1 - x2}[/tex]

Second equation

J = [tex]\frac{M}{A*t}[/tex]

To find the J (flux) use the First equation

J = - 1.7 * [tex]10^{8} m^{2}/s[/tex] * [tex]\frac{2 kg/m^{3} - 0.4 kg/m^{3}}{0 - 6 * 10^{-3} } = 4.53 * 10^{-6} \frac{kg}{m^{2}s }[/tex]

To find M use the Second equation

[tex]4.53 * 10^{-6} \frac{kg}{m^{2}s}[/tex] = [tex]\frac{M}{0.25 m^{2} * 3600s/h}[/tex]

M = [tex]4.1 * 10^{-3} \frac{kg}{h}[/tex]

Answer:

[tex]\eta=0.303[/tex]

Explanation:

Given that

Temperature of the hot reservoir ,T₁ = 970 K

Temperature of the cold reservoir ,T₂ = 480 K

We know that only Carnot engine is the ideal engine which gives us the maximum power.The efficiency of Carnot engine is given as

[tex]\eta_{max}=1-\dfrac{T_2}{T_1}[/tex]

[tex]\eta_{max}=1-\dfrac{480}{970}[/tex]

[tex]\eta_{max}=0.505[/tex]

Therefore the efficiency of the given engine will be

[tex]\eta=\dfrac{3}{5}\eta_{max}[/tex]

[tex]\eta=\dfrac{3}{5}\times 0.505[/tex]

[tex]\eta=0.303[/tex]