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
Net force acting on the car is 3 x 10^3 Newtons:
Answer:
Object A will get 200 & object B will get 400 Potential Energy
Explanation:
Potential Energy = mgh (mass x gravity x height)
For Object A :-
2 (mass) x 10 (approx gravity on Earth) x 10 (distance/height)
= 200
For Object B :-
4 x 10 x 10
= 400
Hence Object B will have High Potential Force than A.
Hope it helps. Thanks
Answer:
1.058 decapoise
Explanation:
The specific gravity of a substance is equal to the density of substance in CGS system and if the specific gravity is multiplied by 1000 then we get the density of substance in MKS system.
Density of steel, ρ = 7800 kg/m^3
density of oil, σ = 900 kg/m^3
v = 0.6 m/s
diameter = 13 mm
radius, r = 6.5 mm
Let η be the coefficient of viscosity.
Use the formula for the terminal velocity
η = 1.058 decapoise