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3 years ago

What is the part of the sound wave called where molecules are expanded and are farther apart?

Physics

1 answer:

Lesechka [4]3 years ago

4 0

Answer:

Rarefactions

Explanation:

The sound waves travel in compressions and rarefactions.

As the name suggests compressions are the areas which particles are closer than their usual positioning.

In rarefactions the vibrating particles can be observed further apart than the usual.

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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at

netineya [11]

Answer:

Δu=1300kJ/kg

Explanation:

Energy at the initial state

$p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)$

Is saturated vapor at initial pressure we have

$p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)$

Process 2-3 is a constant volume process

$p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)$

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

$=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg$

Δu=1300kJ/kg

3 0

3 years ago

The four-firm concentration ratio for newstands is 17 and for electric lamp makers it is 89. The HHI for newstands is 105 and f

TEA [102]

Answer:

is; is not

Explanation:

The four-firm concentration ratio for newstands is 17 and for electric lamp makers it is 89. The HHI for newstands is 105 and for electric lamp makers it is 2 comma 850. The newstand market is an example of monopolistic competition. The electric lamp market is not an example of monopolistic competition

4 0

3 years ago

An object starts at the 100 m mark and ends up at the -100 m mark, 50

horsena [70]

It traveled 200 m in 50 seconds. 200/50 can be simplified to 4 m/s!
The velocity is -4 m/s (negative because it travelled from 100 to -100 or backwards)

8 0

2 years ago

A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.

jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

The body is gaining the speed, which means there is a change in kinetic energy.
The change in kinetic energy is equal to the work done.
The friction force is the product of coefficient of the friction and normal force.
The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

7 0

2 years ago

Read 2 more answers