All categories

3 years ago

PLEASE HELP! FOR BRAINLIEST!!!

Physics

2 answers:

Stells [14]3 years ago

7 0

Answer:

I think it's C I'm so sorry if I'm wrong.

Explanation:

SVETLANKA909090 [29]3 years ago

7 0

Answer:

<h2><em>A or</em> <em>Alex > Chai > Jimmy > Sanni</em></h2>

Explanation:

__________________________________________________________

<em>Here's why this is the answer,</em>

<em>Alex's Range = 3.20 - 0.90 = 2.30</em>

<em>Sanni's Range = 1.35 - 1.30 = 0.05</em>

<em>Chai's Range = 2.10 - 1.05 = 1.05</em>

<em>Jimmy's Range = 2.20 - 1.55 = 0.65</em>

__________________________________________________________

<u>Greatest to Least</u>

<em>Alex > Chai > Jimmy > Sanni</em>

<em>Here's why this is correct.</em>

<em>Jimmy = 0.65</em>

<em>Sanni = 0.05</em>

<em>I do not see the answer I have on this question so I think it is A.</em>

<em>Hope this helps! <3</em>

You might be interested in

If two firecrackers produce a sound level of 81 dBdB when fired simultaneously at a certain place, what will be the sound level

Mrac [35]

Answer:

77.96dB

Explanation:

Recall that decibels are a unit of measuring intensity of sound, and depend on the logarithm of the intensity

the intensity, measured in decibels is given by:

I(db)=10log(I/I0)

I is the intensity in MKS units; I0 is the threshold intensity for human hearing (10^-12 W/m^2)

Thus, if the two sounds together have a dB of 81, we know:

81=10log(I/I0)

using the data above, we can find the intensity of the two sounds to be

0.000125 W/m^2

therefore, one firecracker has an intensity half of that, or 0.0000625W/m^2

now use this value to find the dB of one firecracker:

I(dB0=10log(0.0000625/10^-12)=77.96dB

5 0

3 years ago

When two or more forces act on an object at the same time what are they called

Vilka [71]

Unbalanced forces is what they are called

7 0

3 years ago

To determine a waves frequency you must know the??

lutik1710 [3]

Answer: I think, the number of oscillations in a given period of time.

Explanation: Well I guess because in a period time is known as the rate of occurrence of the wave. Hope this helps!

7 0

3 years ago

Read 2 more answers

Nuclear reactions produce huge amounts of energy by transforming tiny amounts of matter. Please select the best answer from the

vitfil [10]

True, nuclear reactions, whether they are fission or fusion depend on generating energy by converting a certain amount of mass by breaking the atom or combining two atoms together.

3 0

4 years ago

Read 2 more answers

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