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irinina [24]
2 years ago
9

A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and t

he volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.10 L ? (The temperature was held constant.)
Physics
1 answer:
Katen [24]2 years ago
5 0

Answer:

0.358g

Explanation:

Density of Helium = 0.179g/L

ρ=m/v

m=ρv

when the volume was 2L

m1= 0.179*2

m1=0.358g

when the volume increased to 4L

m2= 0.179*4

m2=0.716g

gram of helium added = 0.716g-0.358g

=0.358g

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The key to solve this problem is using the equation I = F.Δt = m.Δv, Δv = vf - vi.

The impulse given to the ball with mass 4Kg is 28 N.s. If the ball were already moving at 3 m/s, to calculate its final velocity:

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Answer:

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Explanation:

<u>Horizontal Motion </u>

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If the maximum horizontal distance is known, we can solve the above equation for h:

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