-3.33 m/s^2
The soccer player decelerates from 10 m/s
to 0 m/s. Because he decelerates the acceleration will be a negative
value. Since he takes 3 seconds to decelerate, divide his original
velocity by 3
One major point to be careful of is to keep p
To explain, I will use the equations for kinetic and potential energy:

<h3>Potential energy </h3>
Potential energy is the potential an object has to move due to gravity. An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>. If gravity = 0 or height = 0, there is no potential energy. Example:
An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2). What is the object's gravitational potential energy? <u>(answer: 5*5*9.8 = 245 J</u>)
(gravitational potential energy is potential energy)
<h3>Kinetic energy</h3>
Kinetic energy is the energy of an object has while in motion. An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary). An example:
An object of 5 kg is moving at 5 m/s. What is the object's kinetic energy? (<u>answer: 5*5 = 25 J</u>)
<h3>Kinetic and Potential Energy</h3>
Sometimes, an object can have both kinetic and potential energy. If an object is moving (kinetic energy) and is above the ground (potential), it will have both. To find the total (mechanical) energy, you can add the kinetic and potential energies together. An example:
An object of 5 kg is moving on a 5 meter table at 10 m/s. What is the objects mechanical (total) energy? (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)
Answer:



Explanation:
From the question we are told that
The length of the string is 
The mass of the string is 
The distance from the bridge to the support post 
The tension is 
Generally the frequency is mathematically represented as
![v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ]](https://tex.z-dn.net/?f=v%20%20%3D%20%20%5Cfrac%7Bn%7D%7B2%20%2A%20L%20%7D%20%20%5B%5Csqrt%7B%20%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D%20%5D)
Where n is and integer that defines that overtones
i.e n = 1 is for fundamental frequency
n = 2 first overtone
n =3 second overtone
Also
is the linear density of the string which is mathematically represented as

=> 
=> 
So for n = 1
![v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ]](https://tex.z-dn.net/?f=v_1%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%20%2A%20%200.62%20%7D%20%20%5B%5Csqrt%7B%20%5Cfrac%7B%20540%7D%7B0.003889%7D%20%7D%20%5D)

So for n = 2
![v_2 = \frac{2}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ]](https://tex.z-dn.net/?f=v_2%20%20%3D%20%20%5Cfrac%7B2%7D%7B2%20%2A%20%200.62%20%7D%20%20%5B%5Csqrt%7B%20%5Cfrac%7B%20540%7D%7B0.003889%7D%20%7D%20%5D)

So for n = 3
![v_3 = \frac{3}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ]](https://tex.z-dn.net/?f=v_3%20%20%3D%20%20%5Cfrac%7B3%7D%7B2%20%2A%20%200.62%20%7D%20%20%5B%5Csqrt%7B%20%5Cfrac%7B%20540%7D%7B0.003889%7D%20%7D%20%5D)
