Answer:
The distance of separation is decreased
Explanation:
From Cuolomb's law, we know that the strength of charge is inversely proportional to the distance of separation between the charges. To mean that increasing the distance let's say from 2m to 3 m would mean initial strength getting form 1/4 to 1/9 which is a decrease. The vice versa is true hence the force of repulsion can increase only when we decrease the distance of separation.
V₁(O2) = 6.50<span> L
</span>p₁(O2) = 155 atm
V₂(acetylene) = <span>4.50 L
</span>p₂(acetylene) =?
According to Boyle–Mariotte law (At constant temperature and unchanged amount of gas, the product of pressure and volume is constant) we can compare two gases that have ideal behavior and the law can be usefully expressed as:
V₁/p₁ = V₂/p₂
6.5/155 = 4.5/p₂
0.042 x p₂ = 4.5
p₂ = 107.3 atm
When Janet leaves the platform, she's moving horizontally at 1.92 m/s. We assume that there's no air resistance, and gravity has no effect on horizontal motion. There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.
She's in the air for 1.1 second before she hits the water.
Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform
(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>
