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3 years ago

By how much does the pressure of a gas in a rigid vessel decrease when the temperature drops from 0 ∘C to –1∘C?

1 answer:

7 0

When a problem says a rigid vessel, it means that volume is constant. At constant V, pressure and temperature are indirectly proportional. We calculate as follows:

P1/T1 = P2/T2
P1/P2 = T1/T2
P1/P2 = 273.15 / 272.15
P1/P2 = 1.00

Hope this helps. Have a nice day.

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2 years ago

Long Snorkel: Inhalation of a breath occurs when the muscles surrounding the human lungs move to increase the volume of the lung

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Answer:

h = 0.362 m

Explanation:

The pressure equation with depth is

P₂ = $P_{atm}$ +ρ g h

The gauge pressure is

P2 - $P_{atm}$ = ρ g h

This is the pressure that muscles can create

P₂ - $P_{atm}$ = 3740 Pa

But still the person needs a small pressure for the transfer of gases, so

P₂ - $P_{atm}$ = 3740 - 188 = 3552 Pa

This is the maximum pressure difference, where the person can still breathe,

Let's clear the height

h = 3552 / ρ g

h = 3552 / (1000 9.8)

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This is the maximum depth where the person can still breathe normally.

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2 years ago

What type of system would allow light to enter and exit, but would keep any

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An isolated system , it does not allow any matter or energy to be exchanged

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3 years ago

Read 2 more answers

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The Drag Force equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (air in this case)

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$ , $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude the Drag force is four times greater when the speed is doubled.

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3 years ago

In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a

Ghella [55]

Answer:

m = B²qR² / 2 V

Explanation:

If v be the velocity after acceleration under potential difference of V

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Bqv = mv² / R

v = BqR / m

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from (1) and (2)

B²q²R² / m² = 2 Vq / m

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2 years ago