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DIA [1.3K]
3 years ago
8

A student performs a reaction twice. In the second trial, he increases the temperature of the reaction and notes that the reacti

on happens more quickly. The student concludes that this means the reaction was endothermic. What is wrong with the student's reasoning?
Higher temperature increases reaction rates for both endothermic and exothermic reactions.
An endothermic reaction would not speed up at higher temperatures.
Temperature does not affect the rate of a reaction.
The higher temperature in the second trial actually lowered the activation energy of the reaction.
Physics
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer: higher temperature increases reaction rates for both endothermic and exothermic reactions

Explanation: i took the test

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A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
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Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

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An object starts at the 100 m mark and ends up at the -100 m mark, 50
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It traveled 200 m in 50 seconds. 200/50 can be simplified to 4 m/s!
The velocity is -4 m/s (negative because it travelled from 100 to -100 or backwards)
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