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3 years ago

How long will it take the ball to reach the height of 35m?

1 answer:

7 0

Ahhh... hello! I’ve done a science experiment that has asked this sort of question. Not as easy as it sounds... I know. I’ll tell you what I learned from that experiment. The answer will depend on the materials the ball is made of and then also your drop length. After these two things are determined, you can get a pretty close time and measurement. I hope this information helps. Have a great night!
~Brooke❤️

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Why is physics the hardest subject

horrorfan [7]

Physics can demand that you start with a specific result and make general rules. Not reading the text and not solving exercises makes understanding almost impossible. It also involves a lot of math which most people do not like.

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2 years ago

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Two concentric circular loops of wire lie on a tabletop, one inside the other. The inner wire has a diameter of 20.0 cm and carr

Harrizon [31]

Answer:

Explanation:

Magnetic field due to circular wire at the center = μ₀ I / 2 r

I is current and r is radius . μ₀ = 4π x 10⁻⁷.

field B₁ due to inner loop

B₁ = 4π x 10⁻⁷ x 12 / 2 x .20

= 376.8 x 10⁻⁷

Field due to outer loop

B₂ = 4π x 10⁻⁷ x I / 2 x .30

For equilibrium

B₁ = B₂

376.8 x 10⁻⁷ = 4π x 10⁻⁷ x I / 2 x .30

I = 18 A.

The direction should be opposite to that in the inner wire . It should be anti-clockwise.

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3 years ago

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Norma-Jean [14]

Answer:

There are four main types or classes of organic compounds that are found in all living things. These are carbohydrates, lipids, proteins, and nucleic acids. ... All organic compounds contain carbon, usually bonded to hydrogen. Other elements may also be present.

Explanation:

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3 years ago

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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