In a Major scale the half-steps always fall between SUPERTONIC and SUBDOMINANT, and between LEADING TONE and TONIC.
True/False

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:

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:

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:

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:

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:

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]

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) 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:

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:

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 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:

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:

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:

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: 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:

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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