This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
A. 205 cm
This has 3 significant figures: 2,0 and 5.
B. 0.00004 cm
This has 1 significant figure: 4
C. 20 cm
This has 2 significant figures: 2 and 0.
D. 20. cm
This has 2 significant figures: 2 and 0
Answer:
C. the C horizon likely has a rockier texture than the topsoil and subsoil.
Explanation:
because i did it on study island
We can do this with the conservation of momentum. The fact it is elastic means no KE is lost so we don't have to worry about the loss due to sound energy etc.
Firstly, let's calculate the momentum of both objects using p=mv:
Object 1:
p = 0.75 x 8.5 = 6.375 kgm/s
Object 2 (we will make this one negative as it is travelling in the opposite direction):
p = 0.65 x -(7.2) = -4.68 kgm/s
Based on this we know that the momentum is going to be in the direction of object one, and will be 6.375-4.68=1.695 kgm/s
Substituting this into p=mv again:
1.695 = (0.75+0.65) x v
Note I assume here the objects stick together, it doesn't specify - it should!
1.695 = 1.4v
v=1.695/1.4 = 1.2 m/s to the right (to 2sf)
