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storchak [24]
2 years ago
5

If there is 3.4 m3 of methane gas in a container with a pressure of 18.9 atm and the container expands until the methane has a p

ressure of 2.2 atm, what is the final volume of the methane? Temperature is constant at 305 K. A. 12.23 m3 B. 141.37 m3 C. 29.21 m3 OD. 2.53 m3​
Physics
1 answer:
Veseljchak [2.6K]2 years ago
7 0

Answer:

C. 29.21 m³

Explanation:

<u>Given the following data;</u>

  • Initial volume, V1 = 3.4 m³
  • Initial pressure, P1 = 18.9 atm
  • Final pressure, P2 = 2.2 atm

To find the final volume, we would use Boyle's law;

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

PV = K

P_{1}V_{1} = P_{2}V_{2}

Making V2 the subject of formula, we have;

V_{2} = \frac {P_{1}V_{1}}{P_{2}}

Substituting the values into the formula, we have;

V_{2} = \frac {18.9 * 3.4}{2.2}

V_{2} = \frac {64.26}{2.2}

<em>Final volume, V2 = 29.21 m³</em>

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Answer:
3.0883 x 10^10mg

Explanation:
1 kilogram = 1000 000 milligrams
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3 years ago
The angular speed of the hour hand of a clock, in rad/min, is:___________
lapo4ka [179]

The angular speed is defined as:

<h2>                                      ω=\frac{2\pi}{T}</h2>

        where

                                     T=12*60=720min

                                     \omega=\frac{2\pi}{720}

                                     \omega=4.4**10^{-3} rad/min

4 0
3 years ago
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8 0
2 years ago
Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field
Nuetrik [128]

Answer:

q=6.22*10^-10C

Explanation:

Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 80 N/C. Find |q|

E=α/∈, electric field within the plate

α=q/A

A=area of the plate

∈=is the permittivity

substituting , we have

The field magnitude E between them (neglect fringing)

E=q/A∈

q=EA∈

q=0.88*80*8.84*10^-12

q=6.22*10^-10C

3 0
2 years ago
A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?
iris [78.8K]

Answer:

\Delta x=22.67786838m

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2 (1)

Where:

y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis

y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis

v_o=initial\hspace{3}velocity

t=travel\hspace{3}time

g=gravity\hspace{3}constant

\theta=Initial\hspace{3}launch\hspace{3}angle

In this case:

\theta=0

Because the dog jumps horizontally

Let's asume the gravity constant as:

g=9.8

y=0

Because when the dog reach the base the height is 0

y_o=70

v_o=6

Now let's replace the data in (1)

y_o-\frac{1}{2} *(9.8)*t^2+70

Isolating t:

t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473

Finally let's find the horizontal displacement using this equation:

\Delta x=v_o*cos(\theta)*t

Replacing the data:

\Delta x=6*1*3.77964473=22.67786838m

8 0
3 years ago
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