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dezoksy [38]
3 years ago
14

I'm a given chemical reaction, the energy of the products is less than the energy of the reactants. Which statement is true for

this chemical reaction?
A energy is absorbed in the reaction
B energy is released in the reaction
C there is no transfer of energy in the reaction
D energy is lost in the reaction
Physics
2 answers:
garik1379 [7]3 years ago
8 0
The answer would have to be D.
Lady bird [3.3K]3 years ago
5 0

Answer: Option (B) is the correct answer.

Explanation:

A chemical reaction in which there is release of energy into the atmosphere is known as an exothermic reaction.

Also, energy of products is less than the energy of reactants for an exothermic reaction.

For example, A + B \rightarrow C + D + Heat

Whereas a chemical reaction in which heat is absorbed by the reactant molecules is known as an endothermic reaction.

Also, energy of products is more than the energy of reactants in an endothermic reaction.

For example, A + B + Heat \rightarrow C + D

Hence, we can conclude that the statement, energy is released in the reaction is true for this chemical reaction.

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katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

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Answer:

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What force is required to accelerate a 385 kg couch at 0.2 m/s^2 ?
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Answer:

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Explanation:

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6 0
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