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Karolina [17]
2 years ago
13

For the following, determine the final volume in ml (V2) of a gas when it is heated to 373K (T2) when it's starting temperature

was 273K (T1) and its starting volume was 100 ml (V1)?
136.63 ml
136.63 K
73.19 ml
73.19 K

For the previous question, how are volume and temperature related?

temperature and volume are not related
temperature and volume are inversely proportional
temperature and volume are directly proportional
only pressure and volume are related
Physics
1 answer:
Ksju [112]2 years ago
3 0

Answer:

136.63 ml

Explanation:

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Answer:

  a = -7.29 m / s²

Explanation:

For this exercise we must use Newton's second law,

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indicate that the charge of the two spheres is equal

         q₁ = q₂ = q

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Let's reduce the magnitudes to the SI system

        m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg

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In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas
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Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

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Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

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u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

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v² = (2Vv/3) × (9/4)

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The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

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