The acceleration of the runner in the given time is 2.06m/s².
Given the data in the question;
Since the runner begins from rest,
- Initial velocity;
![u = 0](https://tex.z-dn.net/?f=u%20%3D%200)
- Final velocity;
![v = 13.4m/s](https://tex.z-dn.net/?f=v%20%3D%2013.4m%2Fs)
- Time elapsed;
![t = 6.5s](https://tex.z-dn.net/?f=t%20%3D%206.5s)
Acceleration of the runner; ![a = \ ?](https://tex.z-dn.net/?f=a%20%3D%20%5C%20%3F)
<h3>Velocity and Acceleration</h3>
Velocity is the speed at which an object moves in a particular direction.
Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion
![v = u + at](https://tex.z-dn.net/?f=v%20%3D%20u%20%2B%20at)
Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.
To determine the acceleration of the runner, we substitute our given values into the equation above.
![v = u + at\\\\13.4m/s = 0 + (a * 6.5s)\\\\13.4m/s = a * 6.5s\\\\a = \frac{13.4m/s}{6.5s}\\ \\a = 2.06m/s^2](https://tex.z-dn.net/?f=v%20%3D%20u%20%2B%20at%5C%5C%5C%5C13.4m%2Fs%20%3D%200%20%2B%20%28a%20%2A%206.5s%29%5C%5C%5C%5C13.4m%2Fs%20%3D%20a%20%2A%206.5s%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B13.4m%2Fs%7D%7B6.5s%7D%5C%5C%20%5C%5Ca%20%3D%202.06m%2Fs%5E2)
Therefore, the acceleration of the runner in the given time is 2.06m/s².
Learn more about Equations of Motion: brainly.com/question/18486505
Answer:
kidney
Explanation:
thats where your pee goes through
For a Rectangular plate has a voltage of +180V and a 'voltage of -5V. , the second plate has the Electric field mathematically given as
E=21.5*10^3v/m
<h3>What is
the field strength?</h3>
Generally, the equation for the Electric field is mathematically given as
E=v/d
Where
v={180-(-5)}v
v=185v
Therefore
E=185/8.6*10^{-3}
E=21.5*10^3v/m
In conclusion, Electric field
E=21.5*10^3v/m
Read more about electric field
brainly.com/question/9383604
The equation to be used here is the trajectory of a projectile as written below:
y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity
Since the angle is below horizontal, let's use the minus equation. Substituting the values:
- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m
However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.
Answer:
lowest level contains more energy
Explanation: