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serg [7]
3 years ago
13

When jumping, a flea accelerates at an astounding 1000 m/s2 but over the very short distance of 0.50 mm. If a flea jumps straigh

t up, and the air resistance is neglected (a bad approximation in this case), how high does the flea go?
Physics
1 answer:
Nadusha1986 [10]3 years ago
5 0

Answer:

The flea reaches a height of 51 mm.

Explanation:

Hi there!

The equations of height and velocity of the flea are the following:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

While in free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the flea at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration of the flea due to the jump.

v = velocity of the flea at time t.

g = acceleration due to gravity.

First, let's calculate how much time it takes the flea to reach a height of 0.0005 m. With that time, we can calculate the speed reached by the flea during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference on the ground, then, h0 = 0. Since the flea is initially at rest, v0 = 0. Then:

h = 1/2 · a · t²

We have to find the value of t for which h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Now, let's find the velocity reached in that time:

v = v0 + a · t   (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

When the flea is at a height of 0.50 mm, its velocity is 1.00 m/s. This initial velocity will start to decrease due to the downward acceleration of gravity. When the velocity is zero, the flea will be at the maximum height. Using the equation of velocity, let's find the time at which the flea is at the maximum height (v = 0):

v = v0 + g · t

At the maximum height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, let's find the height reached by the flea in that time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

The flea reaches a height of 51 mm.

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we know car is moving in circular path

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put here value

a =  \frac{10^2}{63}

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An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.2 m and charge 0.021 mC. Suppose th
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Answer:

T = 1.12 10⁻⁷ s

Explanation:

This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.

All the charge dq is at a distance r

           dE = k dq / r²

Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry

            cos θ =\frac{dE_x}{dE}

             dEₓ = dE cos θ

              cos θ = x / r

substituting

                dEₓ = k \frac{dq}{r^2 } \ \frac{x}{r}

                DEₓ = k dq x / r³

let's use the Pythagorean theorem to find the distance r

             r² = x² + a²

where a is the radius of the ring

we substitute

              dEₓ = k \frac{x}{(x^2 + a^2 ) ^{3/2} } \ dq

we integrate

               ∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} }  ∫ dq

               Eₓ = k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}

In the exercise indicate that the electron is very central to the center of the ring

                x << a

                Eₓ = k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}

if we expand in a series

                  (\ 1+ (x/a)^2 \  )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2

we keep the first term if x<<a

                 Eₓ = \frac{ k Q}{a^3} \ x

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                 F = q E

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this is a restoring force proportional to the displacement so the movement is simple harmonic,

                 F = m a

                 - \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }

                 \frac{d^2 x}{dt^2} = \frac{keQ}{ma^3}  \ x

the solution is of type

                  x = A cos (wt + Ф)

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                w² = \frac{keQ}{m a^3}

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we substitute

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                T = 2π  \sqrt{\frac{m a^3 }{keQ} }

let's calculate

                T = 2π \sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19}  \ 0.021  \ 10^{-3} }  }

                 T = 2π pi \sqrt{320.426 \ 10^{-18} }

                 T = 2π  17.9 10⁻⁹ s

                 T = 1.12 10⁻⁷ s

6 0
3 years ago
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