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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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o-na [289]
The formula PE(Potential Energy)= mgh

4 0
3 years ago
What is the volume of a marble with a diameter of 3.0
RideAnS [48]

Explanation:

You can solve for volume using radius or diameter.

Sphere Volume   =   4/3 • π • r³     =     ( π •d³)/6

We're given the diameter so let's use that.

Volume = PI * d^3 / 6

Volume = 3.14159 * 3.0^3 / 6

Volume = 3.14159 * 9 / 2

Volume = 14.137 cubic centimeters

7 0
3 years ago
A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates at 1.5m/s2? A. 51.5 N B. 4
const2013 [10]
Use the Second Law of Newton, which states that Net Force = mass times acceleration

F = m * a

F = 50 kg * 1.5 m/s^2 = 75 N

Answer: option C
3 0
3 years ago
As an aid in working this problem, consult Interactive Solution 3.41. A soccer player kicks the ball toward a goal that is 20.0
alekssr [168]

Answer:

V=14.9 m/s

Explanation:

In order to solve this problem, we are going to use the formulas of parabolic motion.

The velocity X-component of the ball is given by:

Vx=V*cos(\alpha)\\Vx=15.7*cos(31^o)=13.5m/s

The motion on the X axis is a constant velocity motion so:

t=\frac{d}{Vx}\\t=\frac{20.0}{13.5}=1.48s

The whole trajectory of the ball takes 1.48 seconds

We know that:

Vy=Voy+(a)*t\\Vy=15.7*sin(31^o)+(-9.8)*(1.48)=-6.42m/s

Knowing the X and Y components of the velocity, we can calculate its magnitude by:

V=\sqrt{Vx^2+Vy^2} \\V=\sqrt{(13.5)^2+(-6.42)^2}=14.9m/s

6 0
4 years ago
A metal object is suspended from a spring scale. The scale reads 920 N when the object is suspended in air, and 750N when the ob
Ostrovityanka [42]

17.3 L.

<h3>Explanation</h3>

The object appears 920 - 750 = 170\;\text{N} lighter in water than in the air. Water has supplied that 170 N of buoyant force.

The size of the buoyant force on an object in water is the same as the weight of water that the object has displaced. The buoyant force on the metal object here is 170 N. The object must have displaced water of the same weight.

g = 9.81 \;\text{N}\cdot\text{kg}^{-1}.

Mass of water displaced:

\text{Mass} = \dfrac{\text{Weight}}{g} = \dfrac{170}{9.81} = 17.3 \;\text{kg}.

Volume of water displaced:

The density of water at room temperature is 1.000\;\text{kg}\cdot\text{dm}^{-3}. Each kilogram of water will occupy a volume of 1 dm³ (one cubic decimeter), which is the same as 1 L (one liter).

V = \dfrac{\text{Mass}}{\text{Density}} =\dfrac{m}{\rho} = \dfrac{17.3\;\text{N}}{1.000\;\text{kg}\cdot\text{dm}^{-3}} = 17.3\;\text{dm}^{-3}=17.3\;\text{L}.

Volume of the object:

The object is completely under water. As a result, the volume of the object will be the same as the volume of water displaced. The volume of the object is also 17.3 L.

3 0
3 years ago
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