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2 years ago

9

She wants to travel back to Madison by train. The distance from Chicago to Madison is 147 miles and takes 2.47 hours. If she mak

es no stops, what is her average speed?

1 answer:

levacccp [35]2 years ago

6 0

Answer:

59.51 mph.

Explanation:

The following data were obtained from the question:

Distance (d) = 147 miles

Time (t) = 2.47 hours.

Speed (S) =?

Speed is defined as the distance travelled per unit time. Mathematically, it is expressed as:

Speed (S) = Distance (d) / time (t)

S = d/t

With the above formula, we can obtain the speed of the girl as illustrated below:

Distance (d) = 147 miles

Time (t) = 2.47 hours.

Speed (S) =?

S = d/t

S = 147 miles / 2.47 hours.

S = 59.51 miles per hour (mph)

Thus, the speed of the girl is 59.51 mph.

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Answers:

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

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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

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$T$ 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)

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

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

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

Ordering this acceerations from largest to smallest:

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

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