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3 years ago

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A gnat takes off from one end of a pencil and flies around erratically for 41.641.6 s before landing on the other end of the sam

e pencil. If the gnat flew a total distance of 6.656.65 m, and the pencil is 0.04630.0463 m long, find the gnat's average speed and the magnitude of the gnat's average velocity.

1 answer:

wlad13 [49]3 years ago

6 0

Answer:

average speed is 0.159 m/s

average velocity = 0.0011 m/s

Explanation:

given data

time = 41.6 s

total distance = 6.65 m

length = 0.0463 m

to find out

average speed and the magnitude average velocity

solution

we know average speed formula is

average speed = $\frac{total distance}{total time}$ ...............1

put here value

average speed = $\frac{6.65}{41.6}$

average speed is 0.159 m/s

and

average velocity formula is

average velocity = $\frac{displacement}{total time}$ ...............2

here displacement is initial point to final point and here is 0.0463 m

put here value

average velocity = $\frac{0.0463}{41.6}$

average velocity = 0.0011 m/s

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3 years ago

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Answer:

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Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
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$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

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Solve using the quadratic formula given that $x \ge 0$ :

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(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

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