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Yuliya22 [10]
3 years ago
8

Two sealed 1 l containers full of gas are at room temperature. Container a has a pressure of 4 atm and container b has a pressur

e of 2 atm. What must be true about the number densities of the gases in these containers?
Physics
1 answer:
loris [4]3 years ago
5 0

Answer:

The number density of the gas in container A is twice the number density of the gas in container B.

Explanation:

Here we have

P·V =n·R·T

n = P·V/(RT)

Therefore since V₁ = V₂ and T₁ = T₂

n₁ = P₁V₁/(RT₁)

n₂ = P₂V₂/(RT₂)

P₁ = 4 atm

P₂ = 2 atm

n₁ = 4V₁/(RT₁)

n₂ =2·V₁/(RT₁)

∴ n₁ = 2 × n₂

Therefore, the number of moles in container A is two times that in container B and the number density of the gas in container A is two times the number density in container B.

This can be shown based on the fact that the pressure  of the container is due to the collision of the gas molecules on the walls of the container, with a kinetic energy that is dependent on temperature and mass, and since the temperature is constant, then the mass of container B is twice that of A and therefore, the number density of container A is twice that of B.

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How do you find the speed of an object given its mass and kinetic energy (what is the formula)?
madam [21]
   v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ; 
        
and solve for "v".

______________________________________________________
Explanation:
_____________________________________________________
The formula is:  KE = (½) * (m) * (v²) ;
_____________________________________
  
"Kinetic energy" = (½) * (mass) * (velocity , "squared")
________________________________________________
Note:  Velocity is similar to speed, in that velocity means "speed and direction";  however, if you "square" a negative number, you will get a "positive"; since:  a "negative" multiplied by a "negative" equals a "positive".
____________________________________________
So, we have the formula:
___________________________________
KE = (½) * (m) * (v²) ;  to solve for "(v)" ; velocity, which is very similar to                                          the "speed"; 
___________________________________________________
we arrange the formula ;
__________________________________________________
(KE) = (½) * (m) * (v²) ;  ↔  (½)*(m)* (v²) = (KE) ; 
___________________________________________________

→ We have:  (½)*(m)* (v²) = (KE)  ; we isolate, "m" (mass) on one side of the equation:
______________________________________________________
   
→ We divide each side of the equation by: "[(½)* (m)]" ; 
___________________________________________________
    
           →   [ (½)*(m)*(v²) ] /  [(½)* (m)]  = (KE) / [(½)* (m)]<span> ;
</span>______________________________________________________
 to get: 
______________________________________________________
                           →   v²     =   (KE) / [(½)* (m)]
                     
                           →   v²     = 2 KE / m
_______________________________________________________
Take the "square root" of each side of the equation ;
_______________________________________________________
                          →  √ (v²)  =  √ { 2*(KE) ] / m }
________________________________________________________

                          →     v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"]; 
       
and solve for "v".

______________________________________________________
8 0
3 years ago
A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0
3 years ago
When a car slows down suddenly, passengers in the car tend to move toward the front of the car. What is this due to?
Neko [114]
Hey there Kendrell!

Yes, this is very true, when the car slows down, our bodies will tend to lean forward a little bit, and this is actually due to the "motion of inertia".

Inertia allows for this to happen, this is why in this case, we have this case.

Hope this helps.
~Jurgen


4 0
3 years ago
A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the b
jonny [76]

Answer

given,

vertical speed of stone,v = 12 m/s

height of the cliff = 70 m

a) time taken by the stone to reach at the bottom of the cliff

We know that,

S = u t + 1/2 a t²

- 70 = 12 t - 0.5 x 9.8 t²

4.9 t² - 12 t - 70 = 0  

solving the equation

t = 5.2 s (neglecting the negative value)

b) again using equation of motion

   v = u + a t

   v = 12 - 9.8 x 5.2

  v = -38.96 m/s

ignoring the negative sign

magnitude of velocity is equal to 38.96 m/s

c) total distance travel by the stone

  vertical distance covered by the stone

 v² = u² + 2 g h

 0 = 12² - 2 x 9.8 x h

 h = 7.34 m

to reach the stone to the same level distance travel be doubled.

Total distance travel by the stone

H = h + h + 70

H = 7.34 x 2 + 70

H = 84.7 m.

8 0
2 years ago
In designing a backyard water fountain, a gardener wants to stream of water to exit from the bottom of one tub and land in a sec
mote1985 [20]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description.

From the definition we know that the speed of a body can be described as a function of gravity and height

V = \sqrt{2gh}

V = \sqrt{2*9.8*0.15}

V = 1.714m/s

Then applying the kinematic equation of displacement, the height can be written as

H = \frac{1}{2}gt^2

Re-arrange to find t,

t = \sqrt{2\frac{h}{g}}

t = \sqrt{2\frac{0.5}{9.8}}

t = 0.3194s

Thus the calculation of the displacement would be subject to

x = vt

x =1.714*0.3194

x = 0.547m

Therefore the required distance must be 0.547m

4 0
3 years ago
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