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jeka94
2 years ago
10

A non-rigid balloon is held in a chamber set at 293 K and 1 atm of pressure. The balloon's volume is 0.05 m³. If the pressure in

the room is increased to 5
atm, what is the new volume of the balloon?
Physics
1 answer:
enyata [817]2 years ago
5 0

A non-rigid balloon is held in a chamber set at 293 K and 1 atm of pressure. The balloon's volume is 0.05 m³. If the pressure in the room is increased to 5 atm, then the new volume of the balloon would be 0.01 meter³

<h3>What is an ideal gas?</h3>

It is an imaginary gas for which the volume occupies by it is negligible, this gas does not exist in a practical situation and the concept of an ideal gas is only the theoretical one, the real gases approach the behaviors of an ideal gas at a very high temperature and very low pressure.

By using Boyle's law

P₁V₁=P₂V₂

Where P₁, and V₁, are initial pressure and  volume

and P₂ and  V₂, are the final pressure and volume

By substituting the values in the above equation

P₁ = 1 atm

V₁= 0.05 meter³

P₂ = 5.0 atm

V₂= ? meter³

The only unknown term which we have to find is the final volume of the balloon

1.0 ×0.05=5.0×V₂

V₂ =0.01 meter³

Thus, the new volume of the balloon would be 0.01 meter³

Learn more about ideal gas from here

brainly.com/question/8711877

#SPJ1

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Newton is the SI unit of force . Newton is the name of a British scientist and the name of unit is to honour him. The unit is actually Kg.m/s 2 The unit can be derived by the formula. Take the example of weight. It's formula is W = mg . We know that the unit of mass is kg and gravity is m/s 2 so the unit of weight becomes kg.m/s 2 This unit is known as a Newton. It is always given a capital letter because it is someone's name. Other units that are always capitalised (upper case) are Ampere (Amp), Watt, Volt, Coulomb, Kelvin, Celsius, Fahrenheit, Curie, Roentgen because they are also people's names.

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3 years ago
The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d
Mademuasel [1]

Answer:

28.11m far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

  (1). \: \:\beta =10 log(\dfrac{I}{I_0})

The ratio of the intensities can be written as

$ \frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} } $

\dfrac{I}{I_0} = \dfrac{A_0}{A}.

And since

A = 4\pi r^2

and

A_0 = 4 \pi r_0^2,

\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2}  = \dfrac{r_0^2}{r^2}

meaning

\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.

Putting this into equation (1), we get:

\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

100dB=10 log(\dfrac{r_02}{(5m)^2})

10= log(\dfrac{r_0^2}{25})

by taking both sides to the exponent:  

10^{10}= \dfrac{r_0^2}{25}

r^2 = 25 *10^{10}\\r = 5 *10^5

Now equation (2) becomes

\beta = 10log(\dfrac{25*10^{10}}{r^2})

when the intensity level is 85 dB we have

85 = 10log(\dfrac{25*10^{10}}{r^2})

8.5 = log(\dfrac{25*10^{10}}{r^2})

take both sides to exponents and we get:

10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}

10^{8.5} =\dfrac{25*10^{10}}{r^2}

r^2 = \dfrac{25*10^{10}}{10^{8.5}}

\boxed{r = 28.11m}

Thus, 28.11m far from the speaker the intensity drops to 85 dB.

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