Synthesis and decomposition reactions are b. Synthesis and decomposition reactions are opposites of each other. Synthesis reactions are in the form A + B → AB, while decomposition reactions are in the form AB → A + B.
Synthesis reactions are reactions in which elements or compounds combine to form a larger compound. They are of the form A + B → AB.
Example: Na(s) + Cl (g)→ NaCl(s)
Decomposition reactions are reactions in which a larger compound breaks up into smaller compounds or elements. They are of the form AB → A + B.
Example: CaCO₃(s) → CaO (s) + CO₂ (g)
Synthesis and decomposition reactions are opposite of each other since synthesis reactions, new substances are formed while in decomposition reactions the compound is broken down into into smaller constituents.
So, Synthesis and decomposition reactions are b. Synthesis and decomposition reactions are opposites of each other. Synthesis reactions are in the form A + B → AB, while decomposition reactions are in the form AB → A + B.
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Answer with explanation :
The negative sign means that the potential energy decreases by the movement of the electron.
negative charge at rest in an electric field moves toward the region of an electric field , so that its potential energy will diminish and change into the kinetic energy of motion. The total energy remains constant.
Positive charges will move downhill because of convention. It is to stay in accordance with other potential theories, particularly gravity, where the "charge" is mass, that moves downwards in the gravitational potential field expressed by ϕ(r)=−GM|r|ϕ(r)=−GM|r|. In an electronic system, howbeit, positive charges are fixed in their position within a component (e.g., a wire), therefore instead of the mobile,the negative charges, electrons, move uphill.
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
