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
What’s the question here?
Answer:
false
Explanation:
it cant defined the messy and clean states
A wave with a period of 1⁄3 second has a frequency of D. 3 Hz. To
calculate this we will use the formula that represents the correlation
between a frequency (f) and a time period (T): T = 1/f. Or: f = 1/T. The
unit for the time period is second "s" while the unit for frequency is
Hertz "Hz" (=1/s). We know that T = 1/3 s. That means that f = 1/(1/3s) =
3 1/s = 3 Hz.
Answer:
The velocity of the ball is 3.52 m/s.
Explanation:
A projectile is any object that moves under the influence of gravity and momentum only. Examples are; a thrown ball, a fired bullet, a kicked ball, thrown javelin, etc.
Given that the ball was thrown vertically upward on the top of a skyscraper of height 61.9 m. So that the velocity can be determined by;
u = 
Where: u is the velocity of the object, H is the height and g is the gravitational force on the object. Given that: H = 61.9 m and g = 10 m/
, then;
u = 
= 
u = 3.5185
The velocity of the ball is 3.52 m/s.
