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3 years ago

8

Moustapha jones drives east at 100km/hr for 3 hours then back west at 80km/hr for 1.5 hours. which pair of answers gives his ave

rage speed and average velocity?

1 answer:

olya-2409 [2.1K]3 years ago

7 0

The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.

The total distance cover in east direction is=100*3=300 km

The total distance cover in the west direction=80*1.5=120 km

The total distance covered is =300+120=420 km

And Total displacement of the car is =300-120=180 km

As we know that the average speed is given as

Avg Speed =Total Distance / Total time

=420/4.5=93.33 km/hr

As we know that the average velocity is given as

Avg Speed =Total Displacement/ Total time

=180/4.5=40 km/hr

Therefore, The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.

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Given that

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Now by putting the values

$2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }$

$62831.85=\dfrac{1}{\sqrt{0.002\times C}}$

$\sqrt{0.002\times C=\dfrac{1}{62831.85}$

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<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

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$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

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T = 83 K.

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