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Morgarella [4.7K]
3 years ago
10

What is it asking MathPhys im sorry i dont know but i tried teaching myself physics but i cant

Physics
1 answer:
RSB [31]3 years ago
6 0

Answer:

The puck B remains at the point of collision.

Explanation:

This is an elastic collision, so both momentum and energy are conserved.

The mass of both pucks is m.

The velocity of puck B before the collision is vb.

The velocity of puck A and B after the collision is va' and vb', respectively.

Momentum before = momentum after

m vb = m vb' + m va'

vb = vb' + va'

Energy before = energy after

½ m vb² = ½ m vb'² + ½ m va'²

vb² = vb'² + va'²

Substituting:

(vb' + va')² = vb'² + va'²

vb'² + 2 va' vb' + va'² = vb'² + va'²

2 va' vb' = 0

va' vb' = 0

We know that va' isn't 0, so:

vb' = 0

The puck B remains at the point of collision.

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Answer: Option (c) is correct answer.

Explanation:

The given reaction equation is as follows.

          Fe_{2}O_{3} + HCl \rightarrow FeCl_{3} + H_{2}O

The number of atoms on reactant side are as follows.

  • Fe = 2
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The number of atoms on product side are as follows.

  • Fe = 1
  • O = 1
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Therefore, in order to balance the equation we will balance the number of atoms in both reactant and product side.

Since number of Fe atoms in reactant side is 2, therefore, multiply FeCl_{3} by 2 in the product side.

Number of O atoms in reactant side is 3, therefore, multiply H_{2}O by 3 in the product side.

Now, the number of H atoms on product side is 6, therefore, multiply HCl by 6 in the reactant side.

The number of chlorine atoms on both reactant and product side equals 6.

Thus, the equation will become as follows.

             Fe_{2}O_{3} + 6HCl \rightarrow 2FeCl_{3} + 3H_{2}O

Hence, the equation is balanced.

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4 years ago
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Explanation:

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A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s
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Answer:

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Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

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v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

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