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3 years ago

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A motorist travels 406 km, West during a 7.0 hr period. What was the average velocity in (a) km/hr and (b) m/sec?

1 answer:

AVprozaik [17]3 years ago

6 0

Average velocity = (displacement) / (travel time)

Without more details, I just have to assume that the 406 km
is the displacement ... that is, the straight-line distance between
the start point and the end point. If there were any turns, loops,
or backtracks during the trip, then this question couldn't be answered,
because we don't know the displacement.

a). Average velocity = (406 km) / (7.0 hours) = 58 km/hr West.

According to the requirements stated in the question, I still
have to convert this answer to a different unit in order to
rightfully claim the full 5-points bounty.

b). (58 km/hour) x (1,000 m/km) x (1 hour / 3600 sec) = 16-1/9 m/sec .

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Explanation:

From the question we are told that

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The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

So

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

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