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IrinaK [193]
3 years ago
5

2.18 The net potential energy between two adjacent ions, EN, may be represented by the following equation: (1) Calculate the bon

ding 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 Equation (1).
Engineering
1 answer:
goblinko [34]3 years ago
3 0

Answer:

2.18

Explanation:

because ya correc

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notka56 [123]
Ummm did you try to add or subtract and multiply or divide that can get your answer
8 0
2 years ago
Six housing subdivisions within a city area are target for emergency service by a centralized fire station. Where should the new
Marina86 [1]

Answer:

Explanation:

Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F

To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))     where i=1 to N-1 and N=6

C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))

A area of the polygon can be found by the following formulaA=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i}) where i=1 to N-1

A=\frac{1}{2}[ (x_{1}  y_{2} -x_{2}  y_{1})+ (x_{2}  y_{3} -x_{3}  y_{2})+(x_{3}  y_{4} -x_{4}  y_{3})+(x_{4}  y_{5} -x_{5}  y_{4})+(x_{5}  y_{6} -x_{6}  y_{5})]

A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)

A=225.5 miles²

Now putting the value of area in Cx and Cy

C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(x_{2}+x_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(x_{3}+x_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(x_{4}+x_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(x_{5}+x_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{x} =15.36

For Cy

C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(y_{2}+y_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(y_{3}+y_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(y_{4}+y_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(y_{5}+y_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{y} =22.55

So coordinates for the fire station should be (15.36,22.55)

5 0
3 years ago
Integer to Float Conversion All labs must be done during lab time. Each labs worth 10 points The lab can be hand in next day wit
andrew-mc [135]

Answer:

Code explained below

Explanation:

.data

msg1: .asciiz "Please input a temperature in celsius: "

msg2: .asciiz "The temperature in Fahrenheit is: => "

num: .float 0.0

.text

main:

#print the msg1

li $v0, 4

la $a0, msg1

syscall

#read the float value from user

li $v0,6 #read float syscall value is $v0

syscall #read value stored in $f0

#formula for celsius to fahrenheit is

#(temperature(C)* 9/5)+32

#li.s means load immediate float

#copy value 9.0 to $f2

li.s $f2,9.0  

#copy value 5.0 to $f3

li.s $f3,5.0

# following instructions performs: 9/5

#div.s - division of two float numbers

#divide $f2 and f3.Result will stores in $f1

div.s $f1,$f2,$f3

#following instruction performs: temperature(C) * (9/5)

#multiple $f1 and $f0.Result stored in $f1

mul.s $f1,$f1,$f0

#copy value 32 to $f4

li.s $f4,32.0

#following instruction performs: (temperature(C) * (9/5))+32

#add $f1 and $f4.Result stores in $f1

add.s $f1,$f1,$f4

#store float from $f1 to num

s.s $f1,num

#print the msg2

li $v0, 4 #print string syscall value is 4

la $a0, msg2 #copy address of msg2 to $a0

#print the float

syscall

li $v0,2 #print float syscall value is 2

l.s $f12,num #load value in num to $f12

syscall

#terminate the program

li $v0, 10 #terminate the program syscall value is 10

syscall

4 0
3 years ago
An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter
slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg ,  1980.39 MJ/Kmol

Explanation:

molar mass of  ethanol (C2H5OH) = 46 g/mol

molar mass of   octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol =  0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV  of  ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV  of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol)  for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value  ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV  in MJ/kg  = 0.85 * 29.7 + 0.15 * 47.9  = 32.43 MJ/kg

HHV in  MJ /kmol  =  0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0
3 years ago
Which of the following are some of the problems found in city streets?
storchak [24]

Answer: drugs and rushing cars

Explanation: drug dealers are everywhere on city streets nowadays they have been killing young adults

rushing cars or reckless drivers cut curb fast and potentially someone can get hurt they are speeding and not worrying about other people lives at stake

4 0
3 years ago
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