Answer:
232.9m³ (Option b. is the closest answer)
Explanation:
Given:
Air pressure in the lab before the storm, P₁ = 1.1atm
Air volume in the lab before the storm, V₁ = 180m³
Air pressure in the lab during the storm P₂ = 0.85atm
Air volume in the lab before the storm, V₂ = ?
Applying Boyle's law: P₁V₁ = P₂V₂ (at constant temperature)
V₂ = 232.9m³
The air volume in the laboratory that would expand in order to make up for the large pressure difference outside is 232.9m³
A student uses a meter to measure 120 coulombs flowing through a circuit in 60 seconds. The electric current in this circuit will be 2 A
Current is a flow of electrical charge carriers, usually electrons or electron-deficient atoms. The common symbol for current is the uppercase letter I. The standard unit is the ampere, symbolized by A.
current = charge / time
given
time = 60 seconds
charge = 120 Coulombs
current = Q / T = 120 / 60 = 2 A
To learn more about electric current here
brainly.com/question/12791045
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The acceleration of the car will be needed in order to calculate the time. It is important to consider that the final speed is equal to zero:
We can clear time in the speed equation:
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