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
a) A total charge Q = 23.6 μC is deposited uniformly on the surface of a hollow sphere with radius R = 26.1 cm. Use ε0 = 8.85419 X 10−12 C2/Nm2. What is the magnitude of the electric field at the center of the sphere? b) What is the magnitude of the electric field at a distance R/2 from the center of the sphere? c) What is the magnitude of the electric field at a distance 52.2 cm from the center of the sphere?
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
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:
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.
Consider a thin suspended hotplate that measures 0.25 m × 0.25 m. The isothermal plate has a mass of 3.75 kg, a specific heat of 2770 J/kg·K, and a temperature of 250°C. The ambient air temperature is 25°C and the surroundings temperature is 25°C. If the convection coefficient is 6.4 W/m2·K and the emissivity of the plate is 0.42, determine the time rate of change of the plate temperature, , when the plate temperature is 250°C. Evaluate the magnitude of the heat losses by convection and by radiation.
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.
Explanation: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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Create a C language program that can be used to construct any arbitrary Deterministic Finite Automaton corresponding to the FDA definition above. a. Create structs for the: automaton, a state, and a transition. For example, the automaton should have a "states" field, which captures its set of states as a linked list.
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;
}
Two capacitors with capacitances of 16 nF and 24 nF, respectively, are connected in parallel. This combination is then connected to a battery. If the charge on the 16 nF capacitor is 56 nC, what is the charge on the 24 nF capacitor
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.
Explanation: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:
V = Q/C = 56 nC / 16 nFV = 3.5 VSince 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:
Q = CV = 24 nF × 3.5 VQ= 84 nCThe charge on the 24 nF capacitor is 84 nC.
Wheel diameter = 150 mm, and infeed = 0.06 mm in a surface grinding operation. Wheel speed = 1600 m/min, work speed = 0.30 m/s, and crossfeed = 5 mm. The number of active grits per area of wheel surface = 50 grits/cm2. Determine (a) average length per chip, (b) metal removal rate, and (c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work.
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.
In a Major scale the half-steps always fall between SUPERTONIC and SUBDOMINANT, and between LEADING TONE and TONIC.
True/False
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
Explanation: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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A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is replaced by a high-efficiency 75-hp motor that has an efficiency of 96.3 percent. Determine the reduction in the heat gain of the room due to higher efficiency under full-load conditions.
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]
The density of a fluid is given by the empirical equation rho 70:5 exp 8:27 107 P where rho is density (lbm/ft3 ) and P is pressure (lbf/in2 ). (a) What are the units of 70:5 and 8:27 107?
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
A batch chemical reactor achieves a reduction in concentration of compound A from 90 mg/l to 10 mg/L in one hour. if the reaction is known to follow zero order kinetics, determine the value of the rate constant in (mg/L.hr) unit.
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
Consider an aircraft traveling at high speed. At a point on its wing, the local shear stress is 312 N/m2, and the local conductive heat transfer to the wing is 450 kW/m2. Calculate the air velocity and temperature gradients normal to the surface assuming that the surface has a temperature of 330 K.
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: