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Nastasia [14]
3 years ago
12

A gas cylinder holds 0.10 mol of O2 at 150 C and a pressure of 3.0 atm. The gas expands adiabatically until the pressure is halv

ed
Part A
What is the final volume?
Part B
What is the final temperature?
Physics
1 answer:
yKpoI14uk [10]3 years ago
3 0

Answer:

V2 = 1.899*10^-3 m^3

T2 = 347.125 K

Explanation:

Using gas law, we know that

PV = nRT,

Where

V1 = 0.00115743 m^3.

gamma = 1.4

Now, when we solve for final volume, V2 we get

V2 = V1/((P2/P1)^(1/gamma))

V2 = 1.899*10^-3 m^3

Using the same law and method, when we try to solve for the temperature, we find that the final temperature, T2 is

T2 = T1*((V1/V2)^(gamma-1))

T2 = 347.125 K

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If you run at 1.7 m/s FORWARD ,how does this affect the speed of a ball that you throw?
Sever21 [200]

We have a problem about conservation and velocity, we will find that it does affect the speed of the ball, increasing it by 1.7m/s.

There is something called momentum, which we can define as the "quantity of movement", and we can simply write as the product between velocity and mass.

The momentum is conservative, then we have conservation of momentum.

This means that when you run whit the ball in your hands, the momentum of the ball will be equal to your velocity times the mass of the ball, and this must conserve after you throw the ball.

Now with this idea in mind, this means that if you run with a velocity V, and you throw the ball with a velocity V', the velocity of the ball when it leaves your hand will be:

V + V'.

So, if you run with a velocity of 1.7m/s forward and you throw the ball (assuming in the same direction) the speed of the ball will be 1.7m/s larger than if you were to throw it standing still.

If you want to learn more, you can read:

brainly.com/question/13639113

8 0
3 years ago
An air track car with a mass of 6 kg and velocity of 4 m/s to the right collides with a 3 kg car moving to the left with a veloc
sammy [17]

Remember that moment before collision is equal to the moment after collision.

(m1 \times u1) + (m2 \times u2) = (m1 \times v1) + (m2  \times u2)

Plugging in our values,

(6 \times 4) - (3 \times 2) = (6 \times 1) + (3 \times v2) \\ 24 - 6 = 6 \times 3v2 \\ 18 = 18v2 \\ v2 = 1 {ms}^{ - 1}

8 0
3 years ago
Help with this plz!!!
Alex

The answer is:

It is virtual and behind the mirror.

Mirrors produce images, either virtual or real.

Virtual images are produced when the image is behind the mirror. Real images are formed when they image is produced in front of the mirror.

So look at the image produced. It's small and found in behind the mirror.  This means that it is a virtual image. How do I know that its behind the mirror?

When you look at the diagram, the front side of the mirror is the blue side. This is a convex mirror.

6 0
3 years ago
How much positive and negative charges is there in a cup of water?
Rina8888 [55]
In a cup of water it is a positive and negative charge of zero
3 0
3 years ago
An electron is isolated from an atom and exists in vacuum. A group of scientists collectively state that they can remove part of
lesya692 [45]

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Explanation:

The electric charge of an object is a property of the object that is related to the ability of the object to experience/exert an electric force: if the object is electrically charge, then it is attracted or repelled by other electrically charged object.

The electric charge of an object depends on the amount of charged particles it has on it. In particular, the fundamental particles that carry electric charge are:

  • Protons: they carry electric charge of +e
  • Electrons: they carry electric charge of -e

Where "e" is the fundamental charge (e=1.6\cdot 10^{-19}C). Therefore, one proton carry a charge of +e and one electron carry a charge of -e.

An electron is a fundamental particle: this means that it cannot be divided into smaller particles. This also means that it is not possible to remove part of the charge of the electron: in fact, it is said that electric charge exists only as discrete values, being a multiple of e. Therefore, the correct statement is

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Learn more about particles:

brainly.com/question/2757829

#LearnwithBrainly

5 0
3 years ago
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