treetewew
contestada

Mastering Problems
57. Ranking Task Rank the following objects according
to the magnitude of the acceleration, from least to
greatest. Specifically indicate any ties,
A. A falling acorn accelerates from 0.50 m/s to
10.3 m/s in 1.0 s.
B. A car accelerates from 20 m/s to rest in 1.0 s.
C. A centipede accelerates from 0.40 cm/s to 2.0 cm/s
in 0.50 s.
D. While
being hit, a golf ball accelerates from rest to
4.3 m/s in 0.40 s.
E. A jogger accelerates from 2.0 m/s to 1.0 m/s in 8.3 s.

Respuesta :

Answer:

C , E , A , D , B

Explanation:

We evaluate the accelerations for each case, using the formula: a = (vf - vi) / t

A)  a = (10.3 - 0.5 ) / 1 = 9.8 m/s^2  --> magnitude: 9.8 m/s^2

B)  a = (0 - 20) / 1 = - 20 m/s^2  -->  magnitude : 20 m/s^2

C)  a = (0.02 - 0.004) / 1 = 0.016 m/s^2 -->  magnitude : 0.016  m/s^2

D)  a = (4.3 - 0) / 0.4 = 10.75 m/s^2  -->  magnitude : 10.75 m/s^2

E)  a = (1 - 2) / 8.3 = - 0.12 m/s^2  -->  magnitude: 0.12 m/s^2

Then, comparing magnitudes from least to greatest:

C , E , A , D , B

The order of the magnitude of the acceleration, from least to greatest is:

C < E < A < D < B

In physics, acceleration (a) is the rate at which velocity changes (Δv) with time (t).

We will calculate the acceleration for each scenario using the following expression.

[tex]a = \frac{v-u}{t}[/tex]

where,

  • u: initial velocity
  • v: final velocity

A. A falling acorn accelerates from 0.50 m/s to 10.3 m/s in 1.0 s.

Data

  • u: 0.50 m/s
  • v: 10.3 m/s
  • t: 1.0 s

[tex]a = \frac{10.3m/s-0.50m/s}{1.0s} = 9.8 m/s^{2}[/tex]

B. A car accelerates from 20 m/s to rest in 1.0 s.

Data

  • u: 20 m/s
  • v: 0 m/s (rest)
  • t: 1.0 s

[tex]a = \frac{0m/s-20m/s}{1.0s} = -20 m/s^{2}\\\\|a| = 20 m/s^{2}[/tex]

C. A centipede accelerates from 0.40 cm/s to 2.0 cm/s in 0.50 s.

Data

  • u: 0.40 cm/s
  • v: 2.0 cm/s
  • t: 0.50 s

[tex]a = \frac{2.0cm/s-0.40cm/s}{0.50s} = 3.2cm/s^{2} \times \frac{1m}{100cm} = 0.032 m/s^{2}[/tex]

D. While being hit, a golf ball accelerates from rest to 4.3 m/s in 0.40 s.

Data

  • u: 0 m/s (rest)
  • v: 4.3 m/s
  • t: 0.40 s

[tex]a = \frac{4.3m/s-0m/s}{0.40s} = 11 m/s^{2}[/tex]

E. A jogger accelerates from 2.0 m/s to 1.0 m/s in 8.3 s.

Data

  • u: 2.0 m/s
  • v: 1.0 m/s
  • t: 8.3 s

[tex]a = \frac{1.0m/s-2.0m/s}{8.3s} = -0.12 m/s^{2}\\\\|a| = 0.12 m/s^{2}[/tex]

The order of the magnitude of the acceleration, from least to greatest is:

C < E < A < D < B

Learn more: https://brainly.com/question/16274121

Q&A Education