Mathematics, 03.07.2019 11:40 PineaPPle663

When jumping, a flea rapidly extends its legs, reaching a takeoff speed of 1.0 m/s over a distance of 0.50 mm . part a what is the flea's acceleration as it extends its legs?

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Answer from: tmrsavage02p7cj16

The flea's acceleration is 1,000 m/s².

Further explanation

The problem we overlook this time is about uniformly accelerated motion.

The list of variables to be considered is as follows.

$\boxed{u \ or \ v_i = initial \ velocity}$

$\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}$

$\boxed{a = acceleration \ (constant)}$

$\boxed{d = distance \ travelled}$

Given that a take-off speed of 1.0 m/s over a distance of 0.50 mm.

The initial speed was zero from rest. It was asked about flea's acceleration when she extends her legs.

The formula we follow for this problem are as follows:

$\boxed{ \ v^2 = u^2 + 2ad \ }$

a = acceleration (in m/s²)u = initial velocity = 0 m/sv = take off speed = 1.0 m/sd = distance travelled = 0.50 mm

Converted 0.50 mm to meters: $\boxed{0.50 \ mm = 0.50 \times 10^{-3} \ m = 5.0 \times 10^{-4} \ m}$

Steps:

We set the formula so that the acceleration becomes the subject.

$\boxed{ \ v^2 = u^2 + 2ad \ }$

$\boxed{ \ v^2 - u^2 = 2ad \ }$

$\boxed{ \ 2ad = v^2 - u^2 \ }$

$\boxed{ \ a = \frac{v^2 - u^2}{2d} \ }$

Substitute all data into equations.

$\boxed{ \ a = \frac{(1.0)^2 - (0)^2}{2(5.0 \times 10^{-4})} \ }$

$\boxed{ \ a = \frac{1}{10 \times 10^{-4}} \ }$

$\boxed{ \ a = \frac{1}{1 \times 10^{-3}} \ }$

$\boxed{ \ a = 1 \times 10^{3}\ = 1,000 \ m/s^2 \ }$

We get the flea's acceleration, i.e., 1,000 m/s².

Learn moreAbout scientific notation linkFinding the mass of substance linkConverting in cubic link

Keywords: flea, when jumping, rapidly extends its legs, take-off speed, initial velocity, distance, acceleration, uniformly accelerated motion, convert

When jumping, a flea rapidly extends its legs, reaching a takeoff speed of 1.0 m/s over a distance o

Answer from: elijahjacksonrp6z2o7

The Flea acceleration as it extends its leg is $\fbox{\begin\\\ 1000\ m/s^{2}\\\end{minispace}}$ .

Further explanation:

Given:

The takeoff speed of the flea is $1.0\ \text{m/s}$ .

The distance cover by flea is $0.50\ \text{m/s}$ .

To find:

The acceleration of the flea as it extended it legs.

Formula used:

The acceleration of the flea can be obtained by the given formula.

$\fbox{\begin\\\ \math v^{2}=u^{2}+2as\\\end{minisapce}}$ ....(1)

Here, is the final velocity of flea, is the initial velocity of flea, is the acceleration of the flea, is the distance cover by the flea.

Calculation:

The initial velocity of flea is .

The final velocity of flea is $1.0\text{m/s}$ .

The distance cover by flea is in millimeter.

Convert millimeter into meter as follows.

$\fbox{\begin\\\ \begin{aligned}0.50 \text{ mm}&=\dfrac{0.50}{1000}\text{ m}\\&=0.0005\text{ m}\end{aligned}\\\end{minispace}}$

Now substitute for , 1.0 for and 0.0005 for in equation (1) to obtain the value of acceleration.

$\begin{aligned}1^{2}&=0^{2}+(2a\times(0.0005))\\1&=0.001a\\a&=\dfrac{1}{0.001}\\a&=1000\end{aligned}$

Therefore, the acceleration of the flea is as it extends its leg is $\fbox{\begin\\\ 1000\ m/s^{2}\\\end{minispace}}$ .

Learn more:

1. Solution of linear equation: //link

2. Magnitude of the acceleration link

Answer details:

Grade: Middle school

Subject: Mathematics

Chapter: Speed and distance

Keywords: Flea, jump, legs, takeoff speed, acceleration , distance cover, 0.05 mm, 1000 meter per second square. Initial velocity, Final velocity

Answer from: Quest

7.25

step-by-step explanation:

we know that 7 equals to 7.0. we also know that 1/4=0.25. so we can add the two together and we get 7.25

Answer from: Quest

$6.42

step-by-step explanation:

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