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
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