COLLEGE PHYSICS,VOLUME 1
COLLEGE PHYSICS,VOLUME 1
2nd Edition
ISBN: 9781319115104
Author: Freedman
Publisher: MAC HIGHER
Question
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Chapter 27, Problem 50QAP
To determine

(A)

Complete the fission reaction

Expert Solution
Check Mark

Answer to Problem 50QAP

  A95235m+8n01S3890r+L57149a+4n01

Explanation of Solution

Given information:

  A95235m+x0xS3890r+L57149a+4n01

Calculation:

To calculate the no of neutrons release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,
  235+x=90+149+4x=8

So, 8 neutrons will be released from the above equation
Hence, the complete reaction would be
  A95235m+8n01S3890r+L57149a+4n01

To determine

(B)

Complete the fission reaction

Expert Solution
Check Mark

Answer to Problem 50QAP

  P91244a+n01Z40102r+S51131b+12n01

Explanation of Solution

Given information:

  P91244a+n01xyx+S51131b+12n01

Calculation:

The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,
  244+1=x+131+12×1x=102

And, Atomic Number
  91=y+51y=40

So, Check the periodic table and find the element whose atomic number is 40
Hence, the complete reaction would be
  P91244a+n01Z40102r+S51131b+12n01

To determine

(C)

Complete the fission reaction

Expert Solution
Check Mark

Answer to Problem 50QAP

  C96250m+n01S3492e+S62153m+6n01

Explanation of Solution

Given information:

  xyx+n01S3492e+S62153m+6n01

Calculation:

The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,
  x+1=92+153+6×1x=250

And, Atomic Number
  y=34+62y=96

So, Check the periodic table and find the element whose atomic number is 96.

Hence, the complete reaction would be
  C96250m+n01S3492e+S62153m+6n01

To determine

(D)

Complete the fission reaction

Expert Solution
Check Mark

Answer to Problem 50QAP

  F100262m+n01R45112h+C55142s+9n01

Explanation of Solution

Given information:

  F100262m+n01R45112h+xyx+9n01

Calculation:

The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,
  262+1=112+x+9×1x=142

And, Atomic Number
  100=45+yy=55

So, Check the periodic table and find the element whose atomic number is 55.

Hence, the complete reaction would be
  F100262m+n01R45112h+C55142s+9n01

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I do not understand the process to answer the second part of question b. Please help me understand how to get there!
Rank the six combinations of electric charges on the basis of the electric force acting on 91. Define forces pointing to the right as positive and forces pointing to the left as negative. Rank in increasing order by placing the most negative on the left and the most positive on the right. To rank items as equivalent, overlap them. ▸ View Available Hint(s) [most negative 91 = +1nC 92 = +1nC 91 = -1nC 93 = +1nC 92- +1nC 93 = +1nC -1nC 92- -1nC 93- -1nC 91= +1nC 92 = +1nC 93=-1nC 91 +1nC 92=-1nC 93=-1nC 91 = +1nC 2 = −1nC 93 = +1nC The correct ranking cannot be determined. Reset Help most positive
Part A Find the x-component of the electric field at the origin, point O. Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive. ▸ View Available Hint(s) Eoz = Η ΑΣΦ ? N/C Submit Part B Now, assume that charge q2 is negative; q2 = -6 nC, as shown in (Figure 2). What is the x-component of the net electric field at the origin, point O? Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component that points to the right is positive. ▸ View Available Hint(s) Eoz= Η ΑΣΦ ? N/C

Chapter 27 Solutions

COLLEGE PHYSICS,VOLUME 1

Ch. 27 - Prob. 11QAPCh. 27 - Prob. 12QAPCh. 27 - Prob. 13QAPCh. 27 - Prob. 14QAPCh. 27 - Prob. 15QAPCh. 27 - Prob. 16QAPCh. 27 - Prob. 17QAPCh. 27 - Prob. 18QAPCh. 27 - Prob. 19QAPCh. 27 - Prob. 20QAPCh. 27 - Prob. 21QAPCh. 27 - Prob. 22QAPCh. 27 - Prob. 23QAPCh. 27 - Prob. 24QAPCh. 27 - Prob. 25QAPCh. 27 - Prob. 26QAPCh. 27 - Prob. 27QAPCh. 27 - Prob. 28QAPCh. 27 - Prob. 29QAPCh. 27 - Prob. 30QAPCh. 27 - Prob. 31QAPCh. 27 - Prob. 32QAPCh. 27 - Prob. 33QAPCh. 27 - Prob. 34QAPCh. 27 - Prob. 35QAPCh. 27 - Prob. 36QAPCh. 27 - Prob. 37QAPCh. 27 - Prob. 38QAPCh. 27 - Prob. 39QAPCh. 27 - Prob. 40QAPCh. 27 - Prob. 41QAPCh. 27 - Prob. 42QAPCh. 27 - Prob. 43QAPCh. 27 - Prob. 44QAPCh. 27 - Prob. 45QAPCh. 27 - Prob. 46QAPCh. 27 - Prob. 47QAPCh. 27 - Prob. 48QAPCh. 27 - Prob. 49QAPCh. 27 - Prob. 50QAPCh. 27 - Prob. 51QAPCh. 27 - Prob. 52QAPCh. 27 - Prob. 53QAPCh. 27 - Prob. 54QAPCh. 27 - Prob. 55QAPCh. 27 - Prob. 56QAPCh. 27 - Prob. 57QAPCh. 27 - Prob. 58QAPCh. 27 - Prob. 59QAPCh. 27 - Prob. 60QAPCh. 27 - Prob. 61QAPCh. 27 - Prob. 62QAPCh. 27 - Prob. 63QAPCh. 27 - Prob. 64QAPCh. 27 - Prob. 65QAPCh. 27 - Prob. 66QAPCh. 27 - Prob. 67QAPCh. 27 - Prob. 68QAPCh. 27 - Prob. 69QAPCh. 27 - Prob. 70QAPCh. 27 - Prob. 71QAPCh. 27 - Prob. 72QAPCh. 27 - Prob. 73QAPCh. 27 - Prob. 74QAPCh. 27 - Prob. 75QAPCh. 27 - Prob. 76QAPCh. 27 - Prob. 77QAPCh. 27 - Prob. 78QAPCh. 27 - Prob. 79QAPCh. 27 - Prob. 80QAPCh. 27 - Prob. 81QAPCh. 27 - Prob. 82QAPCh. 27 - Prob. 83QAPCh. 27 - Prob. 84QAPCh. 27 - Prob. 85QAPCh. 27 - Prob. 86QAPCh. 27 - Prob. 87QAPCh. 27 - Prob. 88QAPCh. 27 - Prob. 89QAPCh. 27 - Prob. 90QAPCh. 27 - Prob. 91QAPCh. 27 - Prob. 92QAPCh. 27 - Prob. 93QAPCh. 27 - Prob. 94QAPCh. 27 - Prob. 95QAPCh. 27 - Prob. 96QAPCh. 27 - Prob. 97QAP
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