pictured on the right are three-point charges q1 =...
pictured on the right are three-point charges q1 = 21.4 μc, q2 = -30.6 μc, and q3 = 57.3 μc arranged according to the figure on the right. a fourth point charge is located at point a with a charge of qa = 13.5 μc. calculate the magnitude of the net force on the charge at point a. the answer should be in n.
coulomb's law:f=kq1q2r2, where k is the coulomb's constant and r is the separation between the two charges. the magnitude of the net force:fnet=fx2+fy2k = 9 × 109 n·m2/c2the force due to charge q1 is:f1a=kq1qar1a2=(9×109)(21.4×10-6)(13.5×10-6)(80.1×10-2)2[readmore]f1a = 4.05 nthe force due to charge, q2:f2a=kq2qar2a2=(9×109)(30.6×10-6)(13.5×10-6)[(80.1×10-2)2+(80.1×10-2)2]2f2a = 2.9 nforce due to charge, q3:f3a=kq3qar3a2=(9×109)(57.3×10-6)(13.5×10-6)(80.1×10-2)2f3a = 10.9 nthe net force along the x-axis:fx=f1a-f2acos(45°)=4.05-(2.9)cos(45°)fx = 2.0 nthe net force in the y-direction:fy=f3a-f2acos(45°)=10.9-(2.9)cos(45°)fy = 8.85 nthe magnitude of the net force:fnet=fx2+fy2=2.02+8.852fnet = 9.07 nthe magnitude of the net force on the charge at point a is 9.07 n.
?solution 23cq


a) equation for loop abgefa is (35.5)i1 + (6..25 = 21 v
b) equation for loop cbgedc is ( 9.25i2 - ( 6.25i3 = +15 v <p>concepts and reason</p>
<p>the concept used to solve this problem is the formal charge calculation of the oxygen atoms present in the ozone molecule.</p>
<p>fundamentals</p>
<p>a charge that is assigned to an atom, which is present in the molecule, is represented as a formal charge. in two bonded atoms, a formal charge assumes that the electrons are equally shared between them. the formal charge is calculated using the following formula:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>here, \(\mathrm{v}\) is the number of valence electrons, \(\mathrm{n}\) is the number of non-bonding electrons and \(\mathrm{b}\) is the number of bonding electrons.</p>
<p> </p>
<p>the structure is as follows:</p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p><img src="https://img.wizedu.com/images/5af6d1f54822b4dfb8fe89d91ef2011c5fe3020815aa4.png" alt="" width="88" /></p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(\begin{aligned}=& 6-6-\frac{2}{2} \\ &=-1 \end{aligned}\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the marked oxygen atom has 6 valence electrons. it contains 6 non-bonding electrons. the marked oxygen atom is bonded with one oxygen atom. therefore, the bonding electron of oxygen is \(2 .\) substitute the values in a formal charge formula to determine the formal charge of oxygen (marked).</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/39493056a8515708b79350d1976e17b35fe30216f0473.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-2-\frac{6}{2}\)</p>
<p>\(=+1\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the oxygen atom in the middle has 6 valence electrons and has 2 non-bonding electrons. the central oxygen atom is bonded with 1 double bond and 1 single bond. so, there is a total of 3 bonds containing 6 electrons (bonding electrons). the formal charge of the middle oxygen atom can be determined by substituting the values in a formal charge formula.</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/b1122ce7723aa4ab1f9ae5c9e457299e5fe3022ca77a5.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-4-\frac{4}{2}\)</p>
<p>\(=0\)</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p>
<p>the oxygen atom present on the right side contains 6 valence electrons which have 2 non-bonded electrons. the marked oxygen is bonded with 1 double bond with a central oxygen atom. so, the number of bonding electrons is \(4 .\) the formal charge of the corner oxygen atom (right corner) can be determined by substituting the values in the formal charge formula.</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p> hi trinity, let's do this one!newton's famous second law states, in math terms: f = ma applying to the first case, we have: 100 = m(4.00) => m = 100/4 = 25 kgnow apply the formula to the second case: f = 25 x 16 = 400 newtons AgNO3(aq) + NaCl(s) = AgCl(s) + NaNO3  
n(AgCl) = 0.765 g / M(AgCl)  
M(AgCl) = 108 + 35.5 = 143.5 g/mol  
n(AgCl) = 0.765 g /143.5 g/mol = 0.00533 mol  
n(AgCl) = chloride ions = 0.00533 mol  
C(chloride ions) = n/V  
C(chloride ions) = 0.00533 mol/1l = 0.00533 mol/ L (a)  
m(NaCl)=n(NaCl)*M(NaCl)  
n(NaCl) = 0.00533 mol  
M(NaCl) = 23 + 35.5 = 58.5 g/mol  
m(NaCl) = 0.00533 mol * 58.5 g/mol = 0.3115 g (b)  
  
**Answer:**  
**a) 0.00533 mol/ L**  
**b) 0.3115 g** y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923) (a) f = 1/t = 2.11 hz

(b) 2*pi*f = 13.26 radians/s = sqrt(k/m)
m = 0.296 kg
solve for k
(c) (1/2)ka^2 = 0.150 j
solve for amplitude a, in meters let abc be the trianglewhere ab=(a-1)2 cmbc=2√a cmca=(a+1) cmab2=(a-1)2=a2+1-2abc2=(2√a)2=4a2ca2=(a+1)2=a2+1+2ahence ab2+bc2=ac2so Δabc is a right angles triangle at b there are 2 pints in a quart. there are 4 quarts in a gallon. 1 quart = 2 pints The answer to the question is available in the PDF file https://www.assignmentexpert.com/https://www.assignmentexpert.com/homework-answers/physics-answer-3862.pdf Solution


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