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3 years ago

What is striking a match an example of?

2 answers:

6 0

Answer: option B) providing activation energy to a chemical reaction.

Justification:

1) The combustion of a match is a very exothermic energy process.

2) That means that the reaction liberates a high amount of energy.

3) Process that liberate energy are told to be favourable, this is the productos of the reaction are in a more favorable stable position than before reacting. Then why then do not happen spontaeously? It is because the reactions follow a determined path and that path involves the consumption of energy (named activation energy) to form an activated complex that will then lead to the final formation of the products.

4) That means that to start the products must absorb some energy (the activation energy). Striking a match will provide such activation energy required for the match to ignite and the continue burning. Without this energy the match will never start burning.

Phantasy [73]3 years ago

3 0

Answer:

B) providing activation energy to a chemical reaction

Explanation:

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Suppose a distant world with surface gravity of 5.20 m/s2 has an atmospheric pressure of 7.16 ✕ 104 Pa at the surface. (a) What

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Answer:

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Explanation:

p = Pressure on the plate = $7.16\times 10^4\ Pa$

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4 years ago

A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1

GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

$\large{\maltese{\textsf{\underline{To find :-}}}}$

The X and Y coordinates of the rocket relative of firing

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$\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s$

⠀

$\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}$

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The horizontal range of projectile at x.

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$\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}$

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$\large\textsf{\underline{Now substituting the required values}}$

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$\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}$

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The vertical position of projectile at y.

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$\sf \large y = v_{yi} \times t - (\frac{1}{2} \times g \times {t}^{2}) \\ \\ \sf \large y = v_i \times \cos \theta \times t - \frac{1}{2} g {t}^{2}$

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$\textsf{ \large {\underline{Now substituting the required values}} }$

⠀

$\sf y = 100 \times \cos55{ \degree} \times 12 - \frac{1}{2} \times 9.80 \times {12}^{2} \\ \\ \sf y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\ \\ \sf y = 983.04 - 705.6 \\ \\ \underline{ \boxed{ \bold{y = 277.44m}}}$

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<h3>Henceforth, the distance at horizon is 688.32m and at vertical is 277.44m.</h3>

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2 years ago