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Wewaii [24]
4 years ago
11

If a 70kg swimmer pushes off a pool wall with a force of 250N. What is her acceleration?

Physics
2 answers:
alexgriva [62]4 years ago
7 0
Since , F = ma 250 = 70 × a a = 25/7 m/s²
Alenkasestr [34]4 years ago
6 0

Newton taught us:                              Force = (mass) x (acceleration)

Divide each side by (mass) :              Acceleration = (force) / (mass) .

The only problem here is:  This formula applies when the "Force" is the
only force on the object.  When the objects in these school problems are
falling out of airplanes, shot from guns, or being hit by baseball bats, we
routinely ignore the force of air resistance against the object.  We're
comfortable with that, maybe because it's become a habit.  But now,
we're not so comfortable about ignoring the force of water resistance.

All I can tell you is that if you DO ignore the water resistance, that is,
if the water were not there, her acceleration would be

                    (250 newtons) / (70 kg) = 3.57 m/s² = about  0.36 g .

But what is it really, in the water ? 

If you've spent any substantial amount of time anywhere near competitive
swimmers, then you know that it depends on their position coming off the
wall, what they do with their knees and knuckles, how straight they hold
their body, how deep the texture of their swim-cap is, and how well they've
shaved their legs.

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An egg is thrown horizontally off the top row bleachers at the Brickyard,
Ludmilka [50]

Answer:

4

Explanation:

Divide 30 meters by 7.5 and you´re answer is 4. This is how I would think you solve the problem

6 0
3 years ago
A car stops in 130 m. If it has an acceleration of -5 m/s2 what was the cars starting velocity?
Tatiana [17]

Answer:

<u>We are given:</u>

displacement (s) = 130 m

acceleration (a) = -5 m/s²

final velocity (v) = 0 m/s      [the cars 'stops' in 130 m]

initial velocity (u) = u m/s

<u>Solving for initial velocity:</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(0)² - (u)² = 2(-5)(130)

-u² = -1300

u² = 1300

u = √1300

u = 36 m/s

8 0
3 years ago
14. When incoming light rays strike a flat, plane mirror at an
Mariulka [41]

Answer:

close to the mirrors surface

Explanation:

this is because angle of incidence equals angle of reflection

3 0
3 years ago
I hope y'all can read this I need help​
Ivanshal [37]

Answer:

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4 0
3 years ago
A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res
Natalija [7]

Answer:

The time constant is \tau = 17.5 \ s    

Explanation:

From the question we are told that

   The spring constant is  k = 11.5 \  N/m

   The mass  of the ball is  m_b  = 490 \ g  = 0.49 \ kg

   The amplitude of the  oscillation t the beginning is x =  6.70 cm = 0.067 \  m

    The amplitude after time t is  x_t = 2.20 cm = 0.022 \  m

    The number of oscillation is N  = 30

Generally the time taken to attain the second amplitude is mathematically represented as

       t  = N  *  T                                            Here  T is the period of oscillation

         t = N * 2\pi \sqrt{\frac{m}{k} }

=>     t = 30 * 2 * 3.142 *  \sqrt{\frac{ 0.490}{11.5} }

=>     t = 38.88 \  s

Generally the amplitude at time t is mathematically represented as

         x(t) = x e^{-\frac{at}{2m} }

Here a is the damping  constant so

 at  t = 38.88 \  s ,  x_t = 2.20 cm = 0.022 \  m

So  

     0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }

=>  0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }

taking natural log of both sides

=>  ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }    

=>   a = 0.028

Generally the time constant is mathematically represented as

    \tau = \frac{m}{a}      

=> \tau = \frac{0.490}{  0.028}    

=> \tau = 17.5 \ s    

4 0
3 years ago
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