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3 years ago

What are the three subatomic particles of an atom?

Physics

2 answers:

ZanzabumX [31]3 years ago

8 0

Proton, neutron, and electron. The proton is positive, the neutron is neutral, and the election is negative.

Anestetic [448]3 years ago

5 0

Protons, neutrons, and electrons<span> are the three main subatomic particles found in an atom. Protons have a</span>positive<span> (+) </span>charge<span>. An easy way to remember this is to remember that both proton and </span>positive<span> start with the</span>letter<span> "</span>P<span>." Neutrons have no electrical </span>charge<span>.</span>

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Ca+ is an example of a(n)

umka21 [38]

Answer:

Calcium

hope this helped mah dude :>

3 0

3 years ago

After a party the host pours the remnants of several bottles of wine into a jug. He then inserts a cork with a 2.15-cm diameter

pychu [463]

Answer:

3893.99675 N

Explanation:

$F_1$ = 125 N

$A_1$ = $\pi 1.075^2$

$A_2$ = $\pi 6^2$

From Pascal's law

$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_2=\dfrac{F_1\times A_2}{A_1}\\\Rightarrow F_2=\dfrac{125\times 6^2}{1.075^2}\\\Rightarrow F_2=3893.99675\ N$

The force that the liquid exerts on the bottom of the bottle is 3893.99675 N

5 0

3 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:

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

A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s

Likurg_2 [28]

Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)

Explanation: Please see the attachments below

7 0

3 years ago

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(no links or files) What types of waves are shown in this image?

rewona [7]

Answer:

Absorbed and reflected

Explanation:

B is reflected because you can’t see the red arrow which represents a light that bounces off of it and a aired up sword because you can see it win through the object shown in the picture

3 0

3 years ago

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