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3 years ago

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An object in simple hormonic motion starts at Amplitude Ao. Due to dampening, after 30 sec, the object is now at 30% of its orig

inal amplitude. What is the dampening constant of this system? A. 11s B. 25s C. 33s D. 84s E. 55s

1 answer:

Nookie1986 [14]3 years ago

4 0

Answer:

C ) 33 s.

Explanation:

In simple harmonic motion damping of amplitude is logarithmic which is expressed as follows

$A=A_0e^{-\frac{-b}{t}$

A/A₀ = .3 (given ) , b is damping constant and t is time which is 30 s here.

$.3 = e^\frac{-b}{t}$

ln .3 = -b / 30

1.2 = b / 30

b = 36 s 0r 33 s

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4 years ago

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FOR 50 POINTS! -- Your car is parked at the high school. After school you take off and travel for 27 km towards Iowa City. At th

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36km

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

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Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

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3 years ago

un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración

sergejj [24]

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

Vf: Velocidad final
Vo: Velocidad inicial
a: Aceleración
d: Distancia recorrida

En este caso:

Vf: 0 m/s, porque el avión se detiene
Vo: 50 m/s
a: ?
d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

$a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}$

a= - 10.42 m/s²

<u><em>La aceleración necesaria para detener el avión es - 10.42 m/s².</em></u>

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3 years ago

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madam [21]

Till the time car is just adjacent to the bicycle we can say

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$t = \frac{49 - 0}{7} = 7 s$

Time taken by cycle to accelerate

$t = \frac{23 - 0}{15} = 1.53 s$

now the distance moved by cycle in time "t"

$d = \frac{23 + 0}{2}*1.53 + 23(t - 1.53)$

distance moved by car in same time

$d = \frac{7t + 0}{2}(t)$

now make them equal

$3.5t^2 = 17.595 - 35.19 + 23t$

$3.5 t^2 - 23t + 17.595 = 0$

$t = 5.68 s$

so cycle will move ahead of car for t = 5.68 s

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3 years ago

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