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4 years ago

10

A boat takes off from the dock at 2.5 m/s and speeds up at 4.2 m/s2 for 6.0 s. How far has the boat traveled? Round your answer

to the nearest whole number.

2 answers:

Zinaida [17]4 years ago

6 0

It's simple bro
91m is the answer

k0ka [10]4 years ago

3 0

The correct answer is: 91m

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A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.

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A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

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Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

$\omega = 25 rpm$

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

$\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s$

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

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$\omega=0.42 rev/s$

Keeping in mind that $1 rev = 2\pi rad$ , the angular speed can be rewritten as

$\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42$

And re-arranginf eq.(1), we can find the frequency:

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C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

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$\omega$ is the angular speed

v is the tangential speed

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$\omega = 2\pi \cdot 0.42 rad/s$ is the angular speed

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So, its tangential speed is

$v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s$

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The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

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Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

$d=(26.4)(20)=528 cm$

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