So, the time needed before you hear the splash is approximately <u>2.06 s</u>.
<h3>Introduction</h3>
Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:
With the following condition :
- t = interval of the time (s)
- h = height or any other displacement at vertical line (m)
- g = acceleration of the gravity (m/s²)
Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :
With the following condition :
- t = interval of the time (s)
- s = shift or displacement (m)
- v = velocity (m/s)
<h3>Problem Solving</h3>
We know that :
- h = height or any other displacement at vertical line = 19.6 m
- g = acceleration of the gravity = 9.8 m/s²
- v = velocity = 343 m/s
What was asked :
= ... s
Step by step :
- Find the time when the object falls freely until it hits the water. Save value as
- Find the time when the sound propagate through air. Save value as
- Find the total time
<h3>Conclusion</h3>
So, the time needed before you hear the splash is approximately 2.06 s.
In this collision, 1/2 of the initial kinetic energy of the first glider is converted into thermal energy.
<h3>In plain English, what is kinetic energy?</h3>
An object's strength as a result ofstrength an object has as a result or motion is known as kinetic energy. Toorder to accelerate an object, a force must be applied. Applying force requires effort on our part. When the work is done, power is transported to the thing, which causes it to move at the athe new, constant pace.
<h3>What does kinetic energy mean, or what are some instances?</h3>
The motion energy is known as kinetic energy, and it is manifested when a particle, object, or group if particles moves. Any moving object uses kinetic energy, including people walking, baseballs being thrown, food falling from tables, and charged particles in electric fields.
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The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is
<span>dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) </span>
<span>where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. </span>
<span>However, cos(T) = x/sqrt(x^2 + R^2) </span>
<span>so the equation becomes </span>
<span>dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) </span>
<span>dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 </span>
<span>Integrating around the ring you get </span>
<span>E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 </span>
<span>E = (R/2*e0)*x*(x^2 + R^2)^-1.5 </span>
<span>we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} </span>
<span>to find the maxima set this = 0, giving </span>
<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>
<span>mult both side by (x^2 + R^2)^2.5 to get </span>
<span>(x^2 + R^2) - 3*x^2 = 0 </span>
<span>-2*x^2 + R^2 = 0 </span>
<span>-2*x^2 = -R^2 </span>
<span>x = (+/-)R/sqrt(2) </span>
It would impact the size by how far it is in km or light years
