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storchak [24]
3 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]3 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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It holds more weight in the regular water.
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3 years ago
Using a complete sentence state what would most likely happen to the production of oxygen by duckweed plans if the intensity and
jeka94

Answer:

This question will be answered based on general photosynthetic understanding. The answer is:

The production of oxygen would increase

Explanation:

The characteristics of most plant forms is their ability to photosynthesize i.e. use solar energy (from sunlight) to make food (chemical energy). The product of this photosynthetic process is OXYGEN gas, which is released as a waste product via the stomata on their leaves. Note that, photosynthesis cannot occur without LIGHT as it provides the energy needed for the process.

Hence, in the duckweed plant like every other photosynthetic plant, the increase in the intensity and duration of exposure to light means the rate at which photosynthesis occurs will be increased. An increased photosynthetic rate means the synthesis of the products will also be increased i.e. glucose and OXYGEN.

6 0
3 years ago
A satellite of Mars, called Phobos, has an orbital radius of 9.4 ✕ 106 m and a period of 2.8 ✕ 104 s. Assuming the orbit is circ
FromTheMoon [43]

Answer:

6.27*10^{23}kg

Explanation:

assume

M= mass of Mars

m=mass of phobos

r=orbital radius

T=period

we can apply F=ma to this orbital motion (considering the cricular motion laws)

where,

F=\frac{GMm}{r^{2} }  and a=rω^2

where ω=\frac{2\pi }{T} and G is the universal gravitational constant.

G = 6.67 x 10-11 N m2 / kg2

F=ma\\\frac{GMm}{r^{2} }=mr(\frac{2\pi }{T} )^{2}\\  M=\frac{r^{3}}{G}  (\frac{2\pi }{T} )^{2}\\M=\frac{(9.4*10x^{6} )^{3}*(2\pi )^{2} }{(2.8*10^{4}) ^{2} *6.67*10^{-11} } \\M=6.27*10^{23}kg

6 0
3 years ago
a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(
horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.    

Explanation:

We can find the velocity of the skier by energy conservation:

E_{1} = E_{2}

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}    (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²  

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}  

v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s    

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!  

6 0
2 years ago
Fifty (50) grams of vinegar and (2) grams of baking soda are mixed together. A bubbling reaction takes place. The mixture is wei
Ludmilka [50]

Answer:

the gas that escaped from the mixture contained the missing weight

5 0
3 years ago
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