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Elena-2011 [213]

3 years ago

11

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

ately 57.0 m . If the track is completely flat and the race car is traveling at a constant 27.5 m/s (about 62 mph ) around the turn, what is the race car's centripetal (radial) acceleration?
What is the force responsible for the centripetal acceleration in this case?
friction, normal, gravity, or weight?

1 answer:

valkas [14]3 years ago

5 0

Answer:

$a_c = 13.26 m/s^2$

Friction Force

Explanation:

As we know that centripetal force is the product of mass and centripetal acceleration

so we know that

$a_c = \frac{v^2}{R}$

so here we have

$v = 27.5 m/s$

$R = 57 m$

so we have

$a_c = \frac{27.5^2}{57}$

$a_c = 13.26 m/s^2$

This acceleration is given by the force which may be towards the center of the circular path

Here in the above case it is possible due to friction force.

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Given that;

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$P_{B}$ = 150 kPa

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$P_{B}$ $V_{B}$ = $m_{B}$ R $T_{B}$

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$V_{B}$ = 265.188 / 150

$V_{B}$ = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

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Total mass, $m_{f}$ = $m_{A}$ + $m_{B}$ = 4.309 + 3 = 7.309 kg

Total volume $V_{f}$ = $V_{A}$ + $V_{B}$ = 1 + 1.77 = 2.77 m³

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$P_{f}$ = $m_{f}$ R $T_{f}$ / $V_{f}$

given that; final temperature $T_{f}$ = 20°C = 293 K

we substitute

$P_{f}$ = ( 7.309 × 0.287 × 293) / 2.77

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