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katen-ka-za [31]

3 years ago

12

A race car completes the final lap of a race on a 2-kilometer circular track. At the start of the lap, it crosses the line at 60

meters/second. It completes the lap 28.57 seconds later, crossing the line at 85 meters/second. Calculate the displacement and average acceleration of the car along the track over the course of the final lap.
A. Displacement is 0 meters, and acceleration is 0.88 meters/second2.
B. Displacement is 2,000 meters, and acceleration is 1.14 meters/second2.
C. Displacement is 0 meters, and acceleration is 70.0 meters/second2.
D. Displacement is 2,000 meters, and acceleration is 80.0 meters/second2.
E. Displacement is 0 meters, and acceleration is 0.70 meters/second2.

2 answers:

fomenos3 years ago

5 0

Answer:

E. Displacement is 0 meters, and acceleration is 0.70 meters/second2.

Vikki [24]3 years ago

3 0

Displacement is 0 meters, and acceleration is 0.70 meters/second2.

hope this helped you

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Answer:

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Explanation:

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$N = W - Fy$

Keep in mind Weight of the suitcase (W) is equal to the suitcase mass times the acceleration caused by gravity (9.81 $\frac{m}{s^{2}}$ . Furthermore, Fy can be replaced using trigonometry as $Fsin(\theta)$ where θ is the angle above the horizontal. So the formula can be written in this way:

$N = mg -Fsin(\theta)$

We need to find the value of θ so we can find the value of N. We can find it out solving the sum of forces on X-axis replacing <em>Fx</em> for Fcos(θ). The equation will be like this:

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Replacing the value of F we will see θ has a value of 55.15°. Now we can use this angle to find the value of N. Replacing mass, the gravity acceleration and the angle by their respective values, we will have the following:

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Given:

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$v=\dot{V}\div a$

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