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3 years ago

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Like charges repel and unlike charges attract. Coulomb’s law states that the force F of attraction or repulsion between two char

ges, q1 and q2 is given by F= kq1q2/f^2, where k is constant and r is the distance between the positive charges. Suppose you were to graph F as a function of r for two positive charges

2 answers:

Artist 52 [7]3 years ago

8 0

Step-by-step explanation:

The electrical force between two unlike charge is given by :

$F=\dfrac{kq_1q_2}{r^2}$

k is electrostatic constant

q₁ and q₂ are two charges

r is the distance between charges

It is clear that the electric force is inversely proportional to the square of the distance between them.

The graph F as a function of r for two positive charges is shown in the attached figure. It is a hyperbola.

goblinko [34]3 years ago

3 0

Answer:

vertical at r=0; horizontal F=0

Step-by-step explanation: edg 2020

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For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[

Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

$= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

$= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}\\\\$

calculate the sum: $S= s_1+s_2+s_3+s_n\\$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$

$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$

8 0

3 years ago

6 multiplication problems that ar involved in the solving of 325×89

gogolik [260]

325*89.1

Let's decompose this into:

325 = 300 + 25

89.1 = 80 + 9 + 0.1

Then we have the multiplication:

(300 + 25)*(80 + 9 + 0.1)

Let's distribute the multiplication:

300*80 + 300*9 + 300*0.1 + 25*80 + 25*9 + 25*0.1

So now we have 6 multiplications that are a lot easier to solve than the initial one that we had.

Then the list of six multiplications involved in solving this problem are:

300*80 = 24,400

300*9 = 2,700

300*0.1 = 30

25*80 = 2,000

25*9 = 225

25*0.1 = 2.5

Now we add all of those and get:

325*89.1 = 24,400 + 2,700 + 30 + 2,000 + 225 + 2.5 = 28,957.5

4 0

3 years ago

Please Help! I can’t figure this question out! Question is in the pic below..

padilas [110]

The answer is 68!!!!!!!!!

4 0

3 years ago

Conner and Jana are multiplying (3⁵6⁸)(3⁹6¹⁰).

ad-work [718]

<h3>Conner work is correct. Jana work is wrong</h3>

<em><u>Solution:</u></em>

<em><u>Given that,</u></em>

<em><u>Conner and Jana are multiplying:</u></em>

$(3^56^8)(3^96^{10})$

Given Conner's work is:

$(3^56^8)(3^96^{10}) = 3^{5+9}6^{8+10} = 3^{14}6^{18}$

We have to check if this work is correct

Yes, Conner work is correct

From given,

$(3^56^8)(3^96^{10})\\\\3^5 \times 6^8 \times 3^9 \times 6^{10}$

Use the following law of exponent

$a^m \times a^n = a^{m+n}$

Therefore,

$3^5 \times 6^8 \times 3^9 \times 6^{10} = 3^5 \times 3^9 \times 6^8 \times 6^{10} = 3^{5+9} \times 6^{8+10} = 3^{14} \times 6^{18}$

<em><u>Given Jana's work is:</u></em>

$(3^56^8)(3^96^{10}) = 3^{5.9}6^{8.10} = 3^{45}6^{80}$

This is incorrect

The powers of same base has to be added. But here, powers are multiplied which is wrong

7 0

3 years ago

This one too plss!!!

goblinko [34]

3.10
I think, Im very sorry if this is wrong!

8 0

3 years ago

Read 2 more answers