HELP 25 PTS ASAP!!!!!! BRAINLIEST
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Answer:
The answer to the question is
The ratio of the two gas pressures , that is Px to Py = 1/6
Step-by-step explanation:
Let the gases Volumes be V₁ and V₂
Where volume of X = V₁ and
volume of Y = V₂
The volume of Y is half the volume of X
∴ V₂ = × V₁
Let the number of moles be n₁ and n₂ in X and Y respectively
therefore n₂ = 3 × n₁
The pressure of the gas in X is Pₓ and the pressure of the gas in Y is then we have
P₁ × V₁ = n₁ × R × T₁ , and P₂ × V₂ = n₂ × R × T₂
(P₁ × V₁)/(n₁ × T₁) = (P₂ × V₂)/(n₂ × T₂)
but T₁ = T₂
Therefore
(P₁ × V₁)/n₁ = (P₂ × V₂)/n₂. However n₂ = 3 × n₁ and V₂ = × V₁ therefore substituting in the equation we have
(P₁ × V₁)/n₁ = (P₂ × × V₁ )/(3 × n₁) from where
P₁ /P₂ = ( × V₁ × n₁)/(V₁×3 × n₁) =0.5/3 = 1/6
The ratio of = 1/6
Answer:
The component form of the vector that represents the velocity of the airplane is
Step-by-step explanation:
Given;
velocity of the airplane, v = 75 mph
direction of the plane, θ = 9°
The vertical component of the velocity is given by;
The horizontal component of the velocity is given by;
Therefore, the component form of the vector that represents the velocity of the airplane is given as;
Answer:
The height of the ball after 3 secs of dropping is 16 feet.
Step-by-step explanation:
Given:
height from which the ball is dropped = 160 foot
Time = t seconds
Function h(t)=160-16t^2.
To Find:
High will the ball be after 3 seconds = ?
Solution:
Here the time ‘t’ is already given to us as 3 secs.
We also have the relationship between the height and time given to us in the question.
So, to find the height at which the ball will be 3 secs after dropping we have to insert 3 secs in palce of ‘t’ as follows:
h(3)=160-144
h(3)=16
Therefore, the height of the ball after 3 secs of dropping is 16 feet.