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2 years ago

5

How many air molecules are in a 13.0×12.0×10.0 ft room (28.2 L=1 ft3)? Assume atmospheric pressure of 1.00 atm, a room temperatu

re of 20.0 ∘C, and ideal behavior.

1 answer:

aksik [14]2 years ago

4 0

Answer:

1963.93 Moles

Explanation:

-We know the standard conversion ratio for the volume of a mole is $1 Mole=22.4L$

Given volume of rooms as $13.0ft\times12.0ft\times10ft=1560 ft^3$

Convert the volume into liters: $1560\times28.2l=43992l$

#From our conversion ratio above, we get the volume of air molecules in moles as:

$V_m=\frac{43992L}{22.4L}\times 1\ mole\\\\=1963.93\ Moles$

Hence, the volume of air molecules is 1963.93 Moles

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La tensión en newtons necesaria para que una onda transversal cuya longitud de onda es 3.33 cm vibre a razón de 625 ciclos por s

NemiM [27]

Answer:

9.34 N

Explanation:

First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

For the waves in this string we have:

$f=625 Hz$ , since it completes 625 cycles per second

$\lambda=3.33 cm = 0.033 m$ is the wavelength

So the speed of the wave is

$v=(625)(0.0333)=20.6 m/s$

The speed of the waves in a string is related to the tension in the string by

$v=\sqrt{\frac{T}{\mu}}$ (1)

where

T is the tension in the string

$\mu=\frac{m}{L}$ is the linear density

In this problem:

$m=16.5 g = 16.5\cdot 10^{-3} kg$ is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

$T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N$

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3 years ago

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vagabundo [1.1K]

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A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ $ln[\frac{P_{1} }{P_{2} }]$

= <em>mRT</em>₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵<em>pV</em> = <em>mRT</em>]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = Δ<em>U </em>+ <em>W</em>

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3 years ago

An airtight box has a removable lid of area 1.10 10-2 m2 and negligible weight. The box is taken up a mountain where the air pre

andrezito [222]

Answer:

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The magnitude of force required to pull the lid off the box by air pressure.

We know that Pressure, P= Force(F)/Area(A)

Force, F= P×A

Given: A= $1.10\times10^{-2} m^2$

P= $7\times10^{4} Pa$ .

Therefore, F= $7\times10^{4}\times1.10\times10^{-2}$ .

F= $7.7\times10^2 N$

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3 years ago