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Alchen [17]
3 years ago
14

A.

Physics
1 answer:
STatiana [176]3 years ago
5 0

Answer:

  1. s 6xdbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

Explanation:

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solniwko [45]
Answer: Static Electricity
6 0
3 years ago
A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi
Norma-Jean [14]

Answer:

Now e is due to the ring at a

So

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

So

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

8 0
3 years ago
Please help with this problem ASAP!!!
Alika [10]

Answer:

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3 0
4 years ago
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:

<h2>F=ma   (1) </h2>

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:


<h2>P=mg   (2) </h2>

<u>For block 1: </u>

P_{1}=m_{1}g   (3)

P_{1}=1.5kg(9.8\frac{m}{s^{2}})    

<h2>P_{1}=14.7N   (4) </h2>

<u>For block 2: </u>

P_{2}=m_{2}g   (5)

P_{2}=2.4kg(9.8\frac{m}{s^{2}})    

<h2>P_{2}=23.52N      (6) </h2>

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

F_{r}=P_{1}+P_{2}   (7)

<h2>P_{1}+P_{2}=(m_{1}+m_{2})a   (8) </h2>

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 earth exerts a force of 1.00 newtons on an object in free fall. what is the objects mass?
ICE Princess25 [194]
On or near the surface of the Earth, 1 newton is the weight of about 102 grams of mass.

Note that the gravitational force between the object and the Earth is always the
same.  It doesn't matter whether the object is falling, flying, floating, fluttering,
rising, sinking, rolling, sliding, or just laying there. It doesn't change.
3 0
3 years ago
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