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.

Answer:

[tex]v(t)=-\frac{5}{2}\sqrt{t^3}+20\\s(t)=-\sqrt{t^5}+20t[/tex]

[tex]s(t=4)=48\text{ m}[/tex]

Explanation:

In this case acceleration is defined as:

[tex]a(t)=-k\sqrt{t}[/tex] ,

where k is a constant to be found.

To find the expressions for velocity and position as a function of time you must integrate the expression above for acceleration two times.

Initial conditions and boundary conditions are defined with the rest of the data as:

[tex]v(t=0)=20\text{ m/s}\\v(t=4)=0\text{ m/s}\\s(t=0)=0\text{ m}[/tex]

First integration is equal to:

[tex]a'(t)=v(t)=-k\int\sqrt{t}dt=-\frac{2}{3}k\sqrt{t^3}+C_1[/tex]

The boundary condition and initial condition can be used to calculate [tex]k[/tex] and [tex]C_1[/tex]:

[tex]C_1=20\\k=\frac{15}{4}[/tex]

With this expression for velocity is defined as:

[tex]v(t)=-\frac{5}{2}\sqrt{t^3}+20[/tex]

The same can be done to get to expression for position:

[tex]s(t)=-\sqrt{t^5}+20[/tex]

To get the total distance traveled you can integrate the velocity expression from time=0 sec to time=4 sec:

[tex]s_{tot}=\int_0^4(-\frac{5}{2}\sqrt{t^3}+20)dt=48\text{ m}[/tex]

Answer:

Detailed working is shown

Explanation:

The attached file shows a detailed step by step calculation..

The radial and transverse components of the rocket's velocity at an angle of 60 degrees from the x-axis are 325 m/s and 563 m/s, respectively.

Using Physics principles, we know that when a rocket moves along a parabolic path, its velocity can be decomposed into two components: the radial component (the component of velocity directly in line with the radial direction) and the transverse component (the component of velocity perpendicular to the radial direction).

Given that the absolute speed |v| of the rocket is 650 m/s and the angle θ that the velocity makes with respect to x-axis (measured counterclockwise) is 60°, we can use the trigonometric definitions of sine and cosine to compute the radial and transverse components respectively.

Part A:  The radial component of velocity (vr) at θ = 60° can be computed using the formula vr = v * cosθ. So, vr = 650 m/s * cos60° = 325 m/s.

Part B: The transverse component of velocity (vt) at θ = 60° can be computed using the formula vt = v * sinθ. So, vt = 650 m/s * sin60° = 563 m/s.

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Answer:

amount of electric charge transported =  1.13 × [tex]10^{-2}[/tex] C

Explanation:

given data

electric current = 237.0 mA = 0.237 A

time = 8 min = 8 × 60 sec = 480 sec

solution

we get here amount of electric charge transported that is express as

amount of electric charge transported = electric current × time  ...........1

put here value and we get

amount of electric charge transported = 0.237  × 480

amount of electric charge transported = 113.76 C

amount of electric charge transported =  1.13 × [tex]10^{-2}[/tex] C

Answer:

The unit of 70.5 is lbm/ft^3

The unit of 8.27×10^7 is in^2/lbf

Explanation:

The unit of 70.5 has the same unit as density which is lbm/ft^3 because exponential is found of constant values (unitless values)

The unit of 8.27×10^7 (in^2/lbf) is the inverse of the unit of pressure P (lbf/in^2) because the units have to cancel out so a unitless value can be obtained. Exponential is found of figures with no unit

Answer:

False

Explanation: Half-steps are two keys that are adjacent. A major scale have half step between 3 and 4,7 and 8. The fourth note- subdominant, and the second note- SUPERTONIC. 7th note-leading tonic, and first note-tonic.

The student's statement is incorrect. In a Major scale, half-steps always fall between the MEDIAN and the SUBDOMINANT, and the LEADING TONE and TONIC

The statement in your question - 'In a Major scale the half-steps always fall between SUPERTONIC and SUBDOMINANT, and between LEADING TONE and TONIC' is False. In a Major scale, the half steps always occur between the 3rd and 4th steps (i.e., MEDIAN and SUBDOMINANT) and between the 7th and 8th steps (i.e., LEADING TONE and TONIC). The SUPERTONIC is the second step of the scale, not directly involved in the half steps of a Major scale. Therefore, the correct sequence of half steps in a Major scale falls between the MEDIAN and the SUBDOMINANT, and the LEADING TONE and TONIC.

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Answer:

charge on the 24 nF capacitor is 84 nC

Explanation:

given data

capacitance Q1 = 16 nF

capacitance Q2 = 24 nF

charge on the 16 nF capacitor C1 = 56 nC

solution

we get here capacitor in parallel have same voltage

V1 = V2   ...............1

here voltage V1 = [tex]\frac{Q1}{C1}[/tex]

[tex]\frac{Q1}{C1}[/tex] = [tex]\frac{Q2}{C2}[/tex]    .....................2

put here value we get

[tex]\frac{56}{16} = \frac{Q2}{24}[/tex]

Q = 84 nC

so charge on the 24 nF capacitor is 84 nC

The charge on the 24 nF capacitor, when paired in parallel with a 16 nF capacitor connected to the same battery, is determined to be 84 nC.

When two capacitors are connected in parallel and then to a battery, they both will have the same voltage across them. Given that the charge (Q) on a capacitor is equal to the product of its capacitance (C) and the voltage (V), and assuming the 16 nF capacitor has a charge of 56 nC, the charge on the 24 nF capacitor can be determined using the formula Q = CV.

First, find the voltage using the 16 nF capacitor:

Since the capacitors are in parallel, the voltage across the 24 nF capacitor is also 3.5 V. Therefore, the charge on the 24 nF capacitor is:

The charge on the 24 nF capacitor is 84 nC.

Answer:

False

Explanation:

DMZ stand for DeMilitarized Zone(perimeter network).

It is a sub-network(physical or logical) that contains external facing services to untrusted network. Not only that it contains it also exposes this kind of services.

Adding another layer of security is the main purpose of DMZ

DMZ is positioned in between the Internet and private network

Answer:

The angular acceleration is -2.44 rad/s², while the linear acceleration is -14.66 in/s².

Explanation:

First we need to find the time, at the given position. W e are given the position of particle to be:

θ = 35°

Converting it to radians because, the given equation is in radians:

θ = (35°) (π radians/180°)

θ = 0.611 radians

Now, we have the equation:

θ = Cos(2t)

2t = Arc Cos (θ)

2t = Arc Cos (0.611 radians)

t = 0.91/2

t = 0.457 sec

Now, to determine angular acceleration of the particle, we must derivate the equation twice with respect to 't'

Angular Velocity = ω = dθ/dt = -2Sin(2t)

Angular Acceleration = α = -4Cos(2t)

Now, we use the value of t:

α = -4Cos(2 x 0.457)

α = -2.44 rad/s² (negative sign shows decceleration)

Now for linear acceleration, we know that:

a = rα

a = (6 in)(-2.44 rad/s²)

a = -14.66 in/s² (negative sign shows decceleration)

Answer:

(a) E = 0 N/C

(b) E = 0 N/C

(c) E = 7.78 x10^5 N/C

Explanation:

We are given a hollow sphere with following parameters:

Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C

R = radius of sphere = 26.1 cm = 0.261 m

Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²

The formula for the electric field intensity is:

E = (1/4πεo)(Q/r²)

where, r = the distance from center of sphere where the intensity is to be found.

(a)

At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.

E = 0 N/C

(b)

Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).

E = 0 N/C

(c)

Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:

E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]

E = 7.78 x10^5 N/C

Answer:

A. 0.0283 mm3/min

B. 15850.2 gal/min

C. 0.2642 gal/min

D. 1.7 m3/hour

Explanation:

A.

[(1 in)3/min *(25.4mm)3/(1 in)]

= 0.02832 mm3/min

B.

[(1m)3/sec*(264.173gal)/(1m)3]*(60secs)/1min

= 15850.2 gal/min

C.

[(Liter/min)*(0.264172gal/liter)]

=0.2642 gal/min

D.

[(1ft)3/min*(0.3048m)3/(1ft)3*(60mins/1hour)]

=1.7 m3/hour

Below is an attachment that should help.

Answer:

see the explanation

Explanation:

/* C Program to construct Deterministic Finite Automaton */

#include <stdio.h>

#include <DFA.h>

#include <stdlib.h>

#include <math.h>

#include <string.h>

#include <stdbool.h>

struct node{

struct node *initialStateID0;

struct node *presentStateID1;

};

printf("Please enter the total number of states:");

scanf("%d",&count);

//To create the Deterministic Finite Automata

DFA* create_dfa DFA(){

  q=(struct node *)malloc(sizeof(struct node)*count);

  dfa->initialStateID = -1;

  dfa->presentStateID = -1;

  dfa->totalNumOfStates = 0;

  return dfa;

}

//To make the next transition

void NextTransition(DFA* dfa, char c)

{

  int tID;

  for (tID = 0; tID < pPresentState->numOfTransitions; tID++){

       if (pPresentState->transitions[tID].condition(c))

      {

          dfa->presentStateID = pPresentState->transitions[tID].toStateID;

          return;

      }

  }

  dfa->presentStateID = pPresentState->defaultToStateID;

}

//To Add the state to DFA by using number of states

void State_add (DFA* pDFA, DFAState* newState)

{  

  newState->ID = pDFA->numOfStates;

  pDFA->states[pDFA->numOfStates] = newState;

  pDFA->numOfStates++;

}

void transition_Add (DFA* dfa, int fromStateID, int(*condition)(char), int toStateID)

{

  DFAState* state = dfa->states[fromStateID];

  state->transitions[state->numOfTransitions].toStateID = toStateID;

  state->numOfTransitions++;

}

void reset(DFA* dfa)

{

  dfa->presentStateID = dfa->initialStateID;

}

Answer:

change interval is 3.93 sec

clearance interval is 1.477 sec

Explanation:

Given data:

upgrade of intersection 4%

street total width at intersection is 60 ft

vehicle length of approaching traffic = 16 ft

speed of approaching traffic =40 mi/hr

85th percentile speed is calculated as

S_{85} = S +5

S_{ 85} = 40 + 5  = 45 mi/h

15th Percentile speed

[tex]S_{15} =S-5[/tex]

          = 40 - 5 = 35 mi/hr

change in interval is calculated as

[tex]y = t + \frac{1.47 S_{85}}{2a +(64.4\times 0.01 G}[/tex]

t is reaction time is 1.0,

deceleration rate is given as 10 ft/s^2

[tex]y = 1.0 +\frac{1.47\times 45}{2a +(64.4\times 0.01\times 4}[/tex]

y = 3.93 s

clearance interval is calculated as

[tex]a_r = \frac{W+ L}{1.47\times S_{15}}[/tex]

[tex]a_r = \frac{60+16}{1.47\times 35} = 1.477 s[/tex]

Answer:

a)  1.165 × 10⁻⁸ N b)- 1.165 × 10⁻⁸ N

Explanation:

Using Coulomb's law

F(attraction) = [tex]\frac{Z1Z2qelectron}{4piER^2}[/tex]

where

R = sum of the distance between the centers of charges = sum of ionic radii = 0.079 nm + 0.120nm = 0.199 nm = 0.199 × 10⁻⁹ m

Z₁ = valency of Mg²⁺ = 2

Z₂ = valency of F ⁻ = - 1

qelectron = charge on electron =1.062 × 10⁻19 C

E = permitivity of free space = 8.85 × 10 ⁻¹² C²/ Nm²

Fa= (1×2× (1.602 × 10⁻¹⁹)²) / (4× 3.142 × 8.85 × 10⁻¹² × (0.199 × 10⁻⁹)²) = 1.165 × 10⁻⁸ N

b) At equilibrium F of repulsion = - F of attraction = - 1.165 × 10⁻⁸ N

Answer: Electric Field Formula An electric charge produces an electric field, which is a region of space around an electrically charged particle or object in which an electric charge would feel force. The electric field exists at all points in space and can be observed by bringing another charge into the electric field.

Explanation:

Explanation:

Note: For equations refer the attached document!

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

 Eq 1

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

Eq 2

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

 Eq 3

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

 Eq 4

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

 Eq 5

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

 Eq 6

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

 Eq 7

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

 Eq 8

For Wheat stone bridge:

 Eq 9

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

Eq9

Substituting e with respective stress-strain relation

Eq 10

part b

Since, axial strain(1-2v) < hoop strain (2-v). V out axial < V out hoop.

Hence, dV hoop < dV axial.

part c  

Given data:

P = 253313 Pa

D = d + 2t = 0.09013 m

t = 65 um

GF = 2

E = 75 GPa

v = 0.33

Use the data above and compute Vout using Eq3

Eq 11  

Answer:

a. [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2

b.833.3(0.006-x)+112

c. 117 deg C

Explanation:

Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron,

Assumption

Heat conduction is steady state and unidimensional  2. thermal conductivity is constant. Heat supplied is not in the plate

4. we disregard heat loss

Heat flux=heat/area

[tex]\alpha[/tex]/A=800W/160*10^-4

with direction to the surface been in the x direction,

the mathematical expression will be

[tex]\frac{d^2T}{dx^2}[/tex]=0..............1

and [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2

from fourier law, for conductivity

T(L)=T2=112C

b. integrating equation 1 twice we have\dT/dx=c1

T(x)=C1x+C2

C1 and C2 are arbitrary constant

at x=0 the boundary conditions become

-kC1=qo

C1=-(qo/k)

at x=L          

=T(L)=C1L+C2=T2

C2=T2-cL1

C2=T2+qoL/k

Juxtaposing C1 and C2 into the general equation , we have

T(x)=-qo/k+T2+qoL/k=qo(L-k)/k+T2

50000*(0.006-x)/60+112

833.3(0.006-x)+112

c. inner surface plate temperature is

T(0)=833.33(0.006-0)+112 ( using the derivation in answer b)

117 deg C

Answer: a) 3mm

b) 5400mm^3/min

c) 4000000chips/min

Explanation:

Wheel diameter(D) =150mm

Infeed(W)=0.06mm

Wheel speed(V)=1600m/min

Work speed(Vw)=0.3m/s

Cross feed(d)=5mm

Number of active grits per area of wheel surface =50grits/cm^2

Average length per chip(Lc)=?

Metal removal rate(Rmr)=?

Number of chips formed per unit time(nc)=?

a) Lc=(Dd)^0.5

Lc=(150*0.06)^0.5

Lc=3mm

b) Rmr=VwWd

Rmr=(0.3m/s)*(10^3mm/m)*(5mm)*(0.06mm)

Rmr=5400mm^3/min

c) nc=VWc

nc=(1600m/min)(10^3)(5mm)(50grits/cm^2)(10^-2)

nc=4,000,000chips/min.

Answer:

value of the rate constant is 80 mg/L-hr

Explanation:

given data

concentration of compound Cao = 90 mg/l

Ca = 10 mg/L

time = 1 hour

solution

we use here  zero order reaction rate flow

-rA = K Ca

[tex]\frac{-dCa}{dt}[/tex] = K

-d Ca = k dt

now we will integrate it on the both side by Cao to ca we get

[tex]- \int\limits^{Ca}_{Cao} {dCa} \, = K \int\limits^4_0 {dt} \,[/tex]  

solve it we get

cao - Ca = K t

put here value  and we get K

90 - 10 = K 1

K = 80 mg/L-hr

so value of the rate constant is 80 mg/L-hr

Answer:

t1 = t3 = 5 seconds

t2 = 15 seconds

Explanation:

For t = t1

a = 10 m/s^2

v(t) = 10*t

s(t) = 5*t^2

Distance traveled = 5*t1^2

For t = t2

a = 0 m/s^2

v(t) = 10*t1

s(t) = 10*t1*t

Distance traveled = 10*t1*t2

For t = t3

a = -10 m/s^2

v(t) = -10*t

s(t) = - 5t^2

Distance traveled = 5t3^2

Sum of all distances = 5*t1^2 + 10*t1*(t2) + 5t3^2

1000 = 5t1^2 + 10t1t2 + 5t3^2 + 10*t1*t2 ....... Eq 1

Distance traveled in first and last segments are the same:

t1 = t3 ..... Eq 2

Given: t1+t2+t3 = 25  .... Eq 3

Solving Equations simultaneously:

Subs Eq 2 into Eq 3 & Eq 1

1000 = 5t1^2 + 10*t1*t2 + 5*t1^2

100 = t1^2 + t1*t2  ..... Eq 4

2t1 + t2 = 25

t2 = 25 - 2t1  .... Eq 5

Subs Eq 5 into Eq 4

100 =  t1^2 + t1*(25 - 2t1)

t1^2 -25t1 + 100 = 0

Solve for t1

t1 = 5 , 20 Hence, t1 = 5 sec is selected

t1 = t3 = 5 sec

t2 = 15 sec

Answer:

He wore his black suit, another color of shirt (not purple) and shoes

Explanation:

Holmes owns two suits: one black and one tweed.

Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.

So, if he wore a bow tie yesterday, it means he wore his black suit, another color of shirt (not purple) and shoes because the shirt color is not purple

Answer:

4.536hp

Explanation:

The decrease in the heat gain of the room is determined from difference in electrical inputs:

[tex]Q = W_{shaft} (\frac{1}{n_{1} } - \frac{1}{n_{2} })\\Q = (75hp)*(\frac{1}{0.91 } - \frac{1}{0.963 })\\\\Q = 4.536 hp[/tex]

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

Answer

//countStrings Method

public class countStrings {

public int Arraycount(String[] arrray, String target) {

int count = 0;

for(String elem : arrray) {

if (elem.equals(target)) {

count++;

}

}

return count;

}

// Body

public static void main(String args [] ) {

countStrings ccount = new countStrings();

int kount = ccount.Arraycount(new String[]{"Sick", "Health", "Hospital","Strength","Health"}, " Health"

System.out.println(kount);

}

}

The Program above is written in Java programming language.

It's Divided into two parts

The first is the method countStrings

While the second part of the program is the main method for the program execution

Answer:

The air velocity is 1442.3m/s

The temperature gradient is 0.00311K/m

Explanation:

Rate of heat transfer = local conductive heat transfer × area

Rate of heat transfer = force × distance/time (distance/time = velocity)

Therefore, rate of heat transfer = force × velocity

Force (F) × velocity (v) = local conductive heat transfer coefficient (k) × Area (A)

F/A × v = k

Shear stress = F/A = 312N/m^2

k = 450kW/m^2 = 450×1000W/m^2 = 450000W/m^2

312 × v = 450000

v = 450000/312 = 1442.3m/s

Air velocity (v) = 1442.3m/s

Temperature gradient = Temperature (T)/distance (s)

From equations of motion

v^2 = u^2 + 2gs

u = 0m/s, v = 1442.3m/s, g = 9.8m/s^2

1442.3^2 = 2×9.8×s

s = 2080229.29/19.6 = 106134.15m

Temperature gradient = 330K/106134.15m = 0.00311K/m

Answer:

The solution is shown in the images attached with the answer.

Explanation:

Answer:

See explanation below.

Explanation:

If the statement is a tautology is true for all the possible combinations

Part a

[tex] (p \land q) \Rightarrow (p \lor r)[/tex] lets call this condition (1)

[tex](p \land q) [/tex] condition (2) and [tex](p \lor r)[/tex] condition (3)

We can create a table like this one:

p       q     r      (2)       (3)     (1)  

T       T     T      T        T       T

T       T     F      T        T       T        

T       F     T      F        T       T

T       F     F      F        T       T

F       T     T      F        T       T

F       T     F      F        F       T

F       F     T      F        T       T

F       F     F      F        F       T

So as we can see we have a tautology.

Part b

[tex] p \Rightarrow (r \Rightarrow p)[/tex] lt's call this condition 1

And [tex] (r \Rightarrow p)[/tex] condition 2

We can create the following table:

p     r       (2)     (1)

T     T       T       T

T     F       T       T

F     T       F       T

F     F       T       T

So is also a tautology.

Answer:

a diameter of D₂ = 0.183 inches would be required

Explanation:

appyling pascal's law

P applied to the hydraulic jack = P required to lift the rock

F₁*A₁ = F₂*A₂

since A₁= π*D₁²/4 ,  A₂= π*D₂²/4

F₁*π*D₁²/4 = F₂* π*D₂²/4

F₁*D₁²=F₂*D₂²

D₂ = D₁ *√(F₁/F₂)

replacing values

D₂ = D₁ *√(F₁/F₂) =  6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches

Answer:

Heat losses by convection, Qconv = 90W

Heat losses by radiation, Qrad = 5.814W

Explanation:

Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:

1. Radiation

2. Conduction

3. Convection

Convection is defined as the transfer of heat through the actual movement of the molecules.

Qconv = hA(Temp.final - Temp.surr)

Where h = 6.4KW/m2K

A, area of a square = L2

= (0.25)2

= 0.0625m2

Temp.final = 250°C

Temp.surr = 25°C

Q = 64 * 0.0625 * (250 - 25)

= 90W

Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.

Qrad = E*S*(Temp.final4 - Temp.surr4)

Where E = emissivity of the surface

S = boltzmann constant

= 5.6703 x 10-8 W/m2K4

Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)

= 5.814 W

The time rate of change of the hotplate's temperature is 0.0062 K/s. The magnitude of the heat losses by convection and radiation is 64.23 W.

The question is asking for the time rate of change of the plate temperature when it is at 250°C and the magnitude of the heat losses by convection and by radiation.

First, transform the initial plate temperature from Celsius to Kelvin, so 250°C = 523.15 K.

The air temperature is also given in Celsius, which is 25°C = 298.15 K.

Next, we calculate the heat loss due to convection using the formula Q_conv = h * A * (T_plate - T_air), where h is the convection coefficient, A is the surface area of the plate, and T_plate and T_air are the temperatures of the plate and the air, respectively.

Substituting the given values, we get: Q_conv = 6.4 W/m^2.k * 0.25 m * 0.25 m * (523.15 K - 298.15 K) = 1.80 W.

The heat loss due to radiation can be calculated using the Stefan-Boltzmann law: Q_rad = ε * σ * A * (T_plate^4 - T_surrounding^4), where ε is the emissivity, σ is the Stefan-Boltzmann constant (5.67 * 10^-8 W/m^2.K^4), and T_surrounding is the surrounding temperature.

Again plugging in the given values, we get the heat loss due to radiation as Q_rad = 0.42 * 5.67 * 10^-8 W/m^2.K^4 * 0.25 m * 0.25 m * (523.15 K^4 - 298.15 K^4) = 62.43 W.

So, the total heat loss Q = Q_conv + Q_rad = 1.80 W + 62.43 W = 64.23 W.

To find the time rate of change of the temperature, we use the formula: dT/dt = Q / (m*C), where dT/dt is the time rate of change of the plate temperature, m is the mass, and C is the specific heat. Substituting the values, we get: dT/dt = 64.23 W / (3.75 kg * 2770 J/kg.K) = 0.0062 K/s.

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Answer:

kindly find attachment for detailed answer

Explanation

Consider a pond that initially contains 10 million gallons of fresh water. Water containing a chemical pollutant flows into the pond at the rate of 5 million gallons per year (gal/yr), and the mixture in the pond flows out at the same rate. The concentration c=c(t) of the chemical in the incoming water varies periodically with time to the expression c(t) = 2 + sin(2t) grams per gallon (g/gal).

Construct a mathematical model of this flow process and determine the amount of chemical in the pond at any time t. Then, plot the solution using Maple and describe in words the effect of the variation in the incoming chemical.

Answer:

s= 0.1156 * w or

s= 0.115*(q+p) in terms of Top Biomass and Root Biomass

Explanation:

Since s (number of seeds) is proportional to biomass (w), and above ground biomass also increases with total plant biomass.

s α w

s= 26

w= 225

k= constant

Thus s = k * w

s/w= k

26/225= k

0.1156= k

The equation showing the relationship between seeds and plant biomass is:

s= 0.1156 * w

Assume q= Top Biomass, and p= Root Biomass

w= q+p

Our equation now becomes

s= 0.115*(q+p)

Answer:

The question is incomplete, below is the complete question

"The real power delivered by a source to two impedance, Z1=4+j5⁡Ω and Z2=10⁡Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."

answer:

a. 615W, 384.4W

b. 17.4A

Explanation:

To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.

recall that the symbol for admittance is Y and express as

[tex]Y=\frac{1}{Z}[/tex]

Hence for each we have,  

[tex]Y_{1} =1/Zx_{1}\\Y_{1} =\frac{1}{4+j5}\\converting to polar \\ Y_{1} =\frac{1}{6.4\leq 51.3}\\ Y_{1} =(0.16 \leq -51.3)S[/tex]

for the second impedance we have

[tex]Y_{2}=\frac{1}{10}\\Y_{2}=0.1S[/tex]

we also determine the voltage cross the impedance,

P=V^2(Y1 +Y2)

[tex]V=\sqrt{\frac{P}{Y_{1}+Y_{2}}}\\[/tex]

[tex]V=\sqrt{\frac{1000}{0.16+0.1}}\\ V=62v[/tex]

The real power in the impedance is calculated as

[tex]P_{1}=v^{2}G_{1}\\P_{1}=62*62*0.16\\ P_{1}=615W[/tex]

for the second impedance

[tex]P_{2}=v^{2}*G_{2}\\ P_{2}=62*62*0.1\\384.4w[/tex]

b. We determine the equivalent admittance

[tex]Y_{total}=Y_{1}+Y_{2}\\Y_{total}=(0.16\leq -51.3 )+0.1\\Y_{total}=(0.16-j1.0)+0.1\\Y_{total}=0.26-J1.0\\[/tex]

We convert the equivalent admittance back into the polar form

[tex]Y_{total}=0.28\leq -19.65\\[/tex]

the source current flows is

[tex]I_{s}=VY_{total}\\I_{s}=62*0.28\\I_{s}=17.4A[/tex]