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3 years ago

At a constant pressure, 10 L of a gas at 546 K is cooled to 273 K

Physics

1 answer:

lisabon 2012 [21]3 years ago

8 0

Answer:

5 L

Explanation:

Ideal gas law:

PV = nRT

If P, n, and R are constant, then:

n₁R/P₁ = n₂R/P₂

Using ideal gas law, we can rewrite this as:

V₁/T₁ = V₂/T₂

This is known as Charles' law.

Plugging in values:

10 L / 546 K = V / 273 K

V = 5 L

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Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

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The time will be:

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$=\frac{2\times 6}{289.9}$

$=\frac{12}{289.9}$

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3 years ago

TRUE OR FALSE! PLZ HELP

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Answer:

True

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2 years ago

Kelli weighs 440 N, and she is sitting on a playground swing that hangs 0.41 m above the ground. Her mom pulls the swing back an

kifflom [539]

Answer:

Explanation:

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3 years ago

A general definition of media is "methods for communicating information." Please select the best answer from the choices provide

igor_vitrenko [27]

Answer:

A. True.

Explanation:

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3 years ago

An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l

butalik [34]

Answer:

a) f ’’ = f₀ $\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }$ , b) Δf = 2 f₀ $\frac{v}{c}$

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

f ’= fo $\frac{c+v}{c}$

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

f ’’ = f’ $\frac{c}{ c-v}$

where c represents the sound velocity in stationary blood

therefore the received frequency is

f ’’ = f₀ $\frac{c}{c-v}$

let's simplify the expression

f ’’ = f₀ \frac{c+v}{c-v}

f ’’ = f₀ $\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }$

b) At the low speed limit v <c, we can expand the quantity

(1 -x)ⁿ = 1 - x + n (n-1) x² + ...

$( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}$

f ’’ = fo $( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )$

f ’’ = fo $( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )$

leave the linear term

f ’’ = f₀ + f₀ 2 $\frac{v}{c}$

the sound difference

f ’’ -f₀ = 2f₀ v/c

Δf = 2 f₀ $\frac{v}{c}$

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3 years ago