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SpyIntel [72]
3 years ago
15

When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.47 mm .What is the flea's acceleration during the j

ump phase?

Physics
2 answers:
garri49 [273]3 years ago
8 0
We can use kinematics here if we assume a constant acceleration (not realistic, but they want a single value answer, so it's implied). We know final velocity, vf, is 1.0 m/s, and we cover a distance, d, of 0.47mm or 0.00047 m (1m = 1000mm for conversion). We also can assume that the flea's initial velocity, vi, is 0 at the beginning of its jump. Using the equation vf^2 = vi^2 + 2ad, we can solve for our acceleration, a. Like so: a = (vf^2 - vi^2)/2d = (1.0^2 - 0^2)/(2*0.00047) = 1,064 m/s^2, not bad for a flea!
kenny6666 [7]3 years ago
8 0

The flea's acceleration during the jump phase is about 1100 m/s²

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem !

This problem is about Kinematics.

We will solve it in the following way

<u>Given:</u>

initial speed = u = 0 m/s

final speed = v = 1.0 m/s

distance traveled = d = 0.47 mm = 4.7 × 10⁻⁴ m

<u>Unknown:</u>

acceleration = a = ?

<u>Solution:</u>

v^2 = u^2 + 2ad

1^2 = 0^2 + 2a(4.7 \times 10^{-4})

a = 1 \div (9.4 \times 10^{-4})

a = 1063.830 ~ m/s^2

a \approx 1100 ~ m/s^2

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

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Explanation:

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- The block of mass m undergoes simple harmonic motion. With the displacement of x from mean position is given by:

                                     x(t) = A*cos(w*t)

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Hence,

                                   t_1 = 0.25 * 2 * pi * sqrt( m / k )

                                   t_1 = 0.5*pi * sqrt( m / k )    

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