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hoa [83]
3 years ago
5

If a car races around a 2.3 mile oval track 220 times, what is the displacement of the car in meters when it crosses the finish

line.
Physics
1 answer:
Wewaii [24]3 years ago
5 0

Answer: 0 miles.

Explanation:

First let's define what the displacement is:

The displacement is defined as the difference between the final position and the initial position.

So, if for example, you start at your house, then go to a store, and then go back to your house, your total displacement will be zero, because the initial position and the final position are the same.

Then when the car reaches the final line, it will be in the same place that it started (because this is an oval track, so the finish line connects with the starting point)

Then the total displacement is 0miles.

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A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the
Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

           v = u + at

           0 = 19.09 - 9.81 t

            t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

     Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

     So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

              S = ut + 0.5 at²

              H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

    Maximum height of the ball = 18.57 m

6 0
3 years ago
For every action, there is an equal and opposite reaction.
KiRa [710]

Answer:

This is newton's 3rd law

Explanation:

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3 years ago
Need help on this thank you
Semmy [17]

Answer:

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3 years ago
A centrifuge in a forensics laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates 46.0 times be
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Answer:

Explanation:

Given,

initial angular speed, ω = 3,700 rev/min

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final angular speed = 0 rad/s

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Angular acceleration

\alpha = \dfrac{\omega_f^2 - \omega^2}{2\theta}

\alpha = \dfrac{0 - 387.27^2}{2\times 92\ pi}

\alpha = -259.28 rad/s^2

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the correct answer is D i just got it on USATestPrep. Your welcome.

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