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Butoxors [25]
3 years ago
8

One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings,among t

hem velocity, acceleration, position, speed, and displacement. The series of questions in this problem is designed to getyou to try to think of these quantities like a physicist.Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components
Part A Velocity differs from speed in that velocity indicates a particle's __________ of motion.Enter the letter from the list given in the problem introduction that best completes the sentence.
Part B Unlike speed, velocity is a __________ quantity.Enter the letter from the list given in the problem introduction that best completes the sentence.
Part C A vector has, by definition, both __________ and direction.Enter the letter from the list given in the problem introduction that best completes the sentence.
Part D Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known collectively as __________.Enter the letter from the list given in the problem introduction that best completes the sentence. D Part E Speed differs from velocity in the same way that __________ differs from displacement.
Physics
1 answer:
Allisa [31]3 years ago
8 0

Mnanan anaw anwkkww wjakana akkqna akaoaman akamakq

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Plz answer brainliest if right!
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Answer: it is D. it is the only possible answer. use the process of elimination. which answers make sense?

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There are no answer chooses to this question

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3 years ago
A small particle starts from rest from the origin of an xy-coordinate system and travels in the xy-plane. Its acceleration in th
CaHeK987 [17]

Answer:

The x-coordinate of the particle is 24 m.

Explanation:

In order to obtain the x-coordinate of the particle, you have to apply the equations for Two Dimension Motion

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Yf=Yo+Voyt+0.5ayt² (II)

Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.

The particle starts from rest from the origin, therefore:

Vox=Voy=0

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Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:

12=0+(0)t+ 0.5(1.0)t²

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Dividing by 0.5 and extracting thr squareroot both sides:

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Xf=0+0t+0.5(2.0)(2√6)²

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