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4 years ago

Why are electrons in different places and not together?

Physics

1 answer:

aksik [14]4 years ago

6 0

I would say because they have the same negative charge and they repel each other

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Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block

Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

This can be read as: The Net Force $F$ of an object is equal to its mass $m$ multiplied by its acceleration $a$ .

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension $T$ of the string (<u>See figure attached). </u>

We already know<u> $m_{2}$ is greater than $m_{1}$ </u>, this means the weight of the block 2 $P_{2}$ is greater than the weight of the block 1 $P_{1}$ ; therefore <u>the acceleration of the system will be in the direction of $P_{2}$ </u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight $P$ is calculated using the <u>acceleration of gravity</u> $g=9.8\frac{m}{s^{2}}$ acting on the blocks:

<u>For block 1: </u>

$P_{1}=m_{1}g$ (3)

$P_{1}=1.5kg(9.8\frac{m}{s^{2}})$

<u>For block 2: </u>

$P_{2}=m_{2}g$ (5)

$P_{2}=2.4kg(9.8\frac{m}{s^{2}})$

Then, we are going to <u>find the acceleration $a$ of the whole system: </u>

$F_{r}=P_{1}+P_{2}$ (7)

Where the Resulting Force $F_{r}$ is equal to the sum of the weights $P_{1}$ and $P_{2}$ .

In the figure attached, note that $P_{1}$ is in opposite direction to the acceleration $a$ , this means it must <u>have a negative sing</u>; while $P_{2}$ is in the same direction of $a$ .

Here we only have to isolate $a$ from equation (8) and substitute the values according to the conditions of the system:

$-14.7N+23.52N=(1.5kg+2.4kg)a$

$8.82N=(3.9kg)a$

Then:

$a=\frac{8.82N }{3.9kg}$

<h2> $a=2.26\frac{m}{ s^{2}}$ </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension $T$ of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.

Let’s prove it:

For $m_{1}$

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

$F_{r}=T-P_{1}$ (9)

$T-P_{1}=m_{1}a$ (10)

$T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the tension of the string </h2><h2> </h2>

For $m_{2}$

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

$F_{r}=T-P_{2}$ (9)

$T-P_{2}=m_{2}a$ (10)

$T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the same tension of the string </h2>

6 0

3 years ago

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum. Find the refractive index of the medium.

grin007 [14]

<h2>Answer:</h2>

The refractive index is 1.66

<h2>Explanation:</h2>

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum .

Refractive index of medium = speed of light in vacuum / speed of light in medium

RI = 1/0.6 = 5/3 or 1.66

3 0

3 years ago

Read 2 more answers

A gold nucleus is 466 fm (1 fm = 10-15 m) from a proton, which initially is at rest. When the proton is released, it speeds away

Artist 52 [7]

Answer:

$v=6.8\times10^6m/s$

Explanation:

The sum of the kinetic and electric potential energies of the proton when initially released must be equal to their sum at infinity, so we have:

$K_i+U_i=K_f+U_f$

Which, since $K_i=0J$ because initially the proton is at rest, is:

$\frac{kqQ}{d}=\frac{mv^2}{2}+\frac{kqQ}{r_\infty}$

where $k=9\times10^9Nm^2/C^2$ is Coulomb's constant, $q=1.6\times10^{-19}C$ the charge of the proton, $Q=(79)(1.6\times10^{-19})C$ the charge of the gold nucleus, since it has 79 protons, $d=466\times10^{-15}m$ the initial separation between them, $m=1.67\times10^{-27}kg$ the mass of the proton and v its final velocity. $r_\infty$ is very far away, so the final electric potential will be 0J, and we have:

$v=\sqrt{\frac{2kqQ}{md}}=\sqrt{\frac{2(9\times10^9)(1.6\times10^{-19})^2(79)}{(1.67\times10^{-27})(466\times10^{-15})}}m/s=6839409m/s$

8 0

3 years ago

The total volume in milliliters of a glucose-water solution is given by the equation below: V = 1001.93 + 111.5282m + 0.64698m2

Ostrovityanka [42]

Answer:

111.657596

Explanation:

The expression of volume is given by

$V=1001.93+111.5282+0.64698m^2$

Partially differentiating the term we get

$\dfrac{\partial V}{\partial x}=\dfrac{\partial (1001.93+111.5282+0.64698m^2)}{\partial x}\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+2\times 0.64698m\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+1.29396m$

m = 0.100

$\dfrac{\partial V}{\partial x}=111.5282+1.29396\times 0.100\\\Rightarrow \dfrac{\partial V}{\partial x}=111.657596$

The partial molar volume of glucose is 111.657596

5 0

4 years ago

Find the half-life of a radioactive material if after 1 year 99.67% of the initial amount remains. (Round your answer to one dec

Svetradugi [14.3K]

Answer:

209.68 years

Explanation:

Let T be the half life.

t = 1 year

N = 99.67 % of No = 0.9967 No

Use the law of radioactivity

N = No x e ^(- λ t)

Where, λ is decay constant.

λ = 0.6931 / T

So,

0.9967 No = No x e^(- λ t)

0.9967 = e^(- λ t)

e^( λ t) = 1 / 0.9967 = 1.0033

λ t = 3.3 x 10^-3

(0.6931 x 1) / T = 3.3 x 10^-3

T = 209.68 years

Thus, the halflife is 209.68 years.

3 0

3 years ago