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sattari [20]
3 years ago
11

A cylinder contains 12liters of O2 at 20°c and 15atm. The temperature is raised to 35°c and volume is reduced to 8.5L. Calculate

the final pressure. What will be the new pressure if the volume is doubled
Physics
1 answer:
svet-max [94.6K]3 years ago
5 0

<h2>Pressure : 11 atmosphere</h2>

Explanation:

According to gas equation  P₁V₁/T₁ = P₂V₂/T₂

Where P₁ is the initial pressure , V₁ is the initial volume and T₁ is the initial temperature

P₂ is the final pressure , V₂ is the final volume and T₂ is the final temperature

Thus P₁ = 15 atmosphere  , V₁ = 12 liters and T₁ = 273 + 20 = 293 K

P₂ = ? , V₂ = 8.5 liters , T₂ = 273 + 35 = 308 K

From gas equation P₂ = P₁V₁T₂/T₁V₂ = 15 x 12 x 308/ 293 x 8.5

= 22.3 atmosphere .

If the volume is doubled

The gas equation will be P₁V₁ = P₂V₂ ;  because temperature is constant

Here V₂ = 2 V₁  therefore  pressure P₂ = P₁/2 = 22.3/2 = 11.1 atmosphere

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A .005kg projectile leaves a 1500kg launcher with a velocity of 750 m/s. What is the recoil velocity of the projectile
docker41 [41]

Answer:

The recoil velocity of the projectile is 0.0025m/s

Explanation:

Given:

Mass of the projectile =0.005kg

Mass of the launcher = 1500kg

Velocity =  750 m/s.

To Find:

The recoil velocity of the projectile = ?

Solution:

The recoil velocity is the obtained by dividing the "recoil momentum"  by the "mass of the recoil body".  The recoil momentum is equal to the momentum of the other body. The momentum of the other body is equal to it mass times its velocity.

Lets find the recoil momentum,

Recoil momentum = mass of the projectile X velocity

Recoil momentum =0.005 \times 750

Recoil momentum = 3.75

Now Recoil Velocity,

Recoil Velocity = \frac{\text { Recoil Momentum}}{\text {Mass of the launcher}}

Recoil Velocity = \frac{ 3.75}{1500}

Recoil Velocity = 0.0025m/s

7 0
4 years ago
The distance between two parallel wires carrying currents of 10 A and 20 A is 10 cm. Determine the magnitude and direction of th
Marianna [84]

Explanation:

It is given that,

Current in wire 1, I₁ = 10 A

Current in wire 2, I₂ = 20 A

Distance between wires, d = 10 cm = 0.1 m

Force per unit length is given by :

\dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}

\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 10\times 20}{2\pi \times 0.1}

\dfrac{F}{l}=0.0004\ N/m

\dfrac{F}{l}=4\times 10^{-4}\ N/m

So, the magnetic force acting per unit length of the wires 4\times 10^{-4}\ N/m. Since, the current is in same direction. So, the force is attractive in nature.

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3 years ago
Maurice pulls on the end of a spring scale. He lets go of the end and observes the spring snap back into place. What force resto
andreev551 [17]
Elastic is the right answer.
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3 years ago
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What is the gauge pressure (in atm) in a 25.0°C car tire containing 3.81 mol of gas in a 31.5 L volume? (Enter your answer to at
bagirrra123 [75]

Answer:

Explanation:

Given

Temperature of gas T=25^{\circ}C\approx 298\ K

Volume of gas V=31.5\ L\approx 31.5\times 10^{-3}\ m^3

no of moles of gas n=3.81\ mol

Using Ideal gas Equation to find the Pressure of gas

PV=nRT

where P=Absolute Pressure

V=Volume

R=Universal gas constant

T=Temperature

P\times 31.5\times 10^{-3}=3.81\times 8.314\times 298

P=\frac{3.81\times 8.314\times 298}{31.5\times 10^{-3}}

P=299.66\ kPa

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Therefore Gauge pressure is given by

P_{gauge}=P-P_{atm}

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4 0
3 years ago
The radius of an atom of krypton (kr) is about 1.9 å. (a) express this distance in nanometers (nm). nm express this distance in
I am Lyosha [343]
Note:
1 A (armstrong) = 10⁻¹⁰ m
1 nm (nanometer) = 10⁻⁹ m

Given:
Radius of a krypton atom = 1.9 A = 1.9 x 10⁻¹⁰ m

Part (a)
1.9\,A = (1.9\,A)*(10^{-10}\, \frac{m}{A})*( \frac{1}{10^{-8}} \frac{nm}{m}) } =0.019\,nm
Answer: 0.019 nm

Part (b)
The diameter of a krypton atom = 2*1.9A = 3.8 A = 3.8 x 10⁻¹⁰ m.
The number of krypton atoms within a length of 1.0 mm is
\frac{1.0\, mm}{3.8 \time 10^{-10}\, m} = \frac{10^{-3}\, m}{3.8 \times 10^{-10} \,m} =2.632 \times 10^{6}

Answer: About 2.632 x 10⁶ atoms

Part (c)
The radius of a krypton atom is
1.9 A = (1.9 x 10⁻¹⁰ m)*(10² cm/m) = 1.9 x 10⁻⁸ cm
The volume of a krypton atm is
\frac{4 \pi }{3} (1.9 \times 10^{-8} \, cm)^{3} = 2.873 \times 10^{-23} \, cm^{3}

Answer: 2.873 x 10⁻²³ cm³
8 0
3 years ago
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