Explanation:
A wave on a string is described is given by :
![D(x,t)=2\ cm\ sin[(12.57\ rad/m)-(638\ rad/s)t]](https://tex.z-dn.net/?f=D%28x%2Ct%29%3D2%5C%20cm%5C%20sin%5B%2812.57%5C%20rad%2Fm%29-%28638%5C%20rad%2Fs%29t%5D)
The linear density of the string is 5 g/m.
Where
x is in meters and t is in seconds
The general equation of a wave is given by :
(2) The speed of the wave in terms of tension is given by :
Also, 
So, 
T = 12.88 N
(3) The maximum displacement of a point on the string is equal to the amplitude of the wave. So, the maximum displacement is 2 cm.
(4) The maximum speed of a point on the string is given by :
v = 12.76 m/s
Hence, this is the required solution.
The man ran <u>4252.5 meters.</u>
Why?
To solve the problem, we need to divide the exercise into two movements, the first on while the was running at 4.5 m/s for 15 min, and then, while he was slowing down (going up because of the hill).
First movement: Running at 4.5 m/s for 15 min.
We need convert from minutes to seconds,
Now, calculating the distance covered for the first movement, we have:
So, we know that the man covered 4050m for the first movement, it will be our initial position for the second movement.
Second movement: acceleration -0.05m/s^2 (because he's slowing down) for 90 seconds, at 4.5m/s.
Hence, we have that he ran 4252.5 m.
Have a nice day!
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
The area of the sprinkles can be determined through the area of a circle that is pi * r^2 in which the given dimensions above are the radii, r. The second scenarios radius is only half of the original, that is 4 ft. In this case, we can compute the area of the second again. We calculate next the difference of two areas of circles.
