The speed of sound at T=25°C is Vs=346 m/s. So the sound has to reach the cliff and return back to you so the path it needs to travel is s=2*440 m = 880 m.
Since the speed of sound is constant s=Vs*t, and t= s/Vs=880/346=2.54335 s. You will hear the echo after t=2.54335 s after you shouted.
The kinetic energy gained by the air molecules is 0.0437 J
<h3 />
Given:
Mass of a coffee filter, m = 1.5 g
Height from which it is dropped, h = 3 m
Speed at ground, v = 0.7 m/s
Initially, the coffee filter has potential energy. It is given by :

P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m
P = 0.0441 J
Finally, it will have kinetic energy. It is given by :

×
× 10⁻³ × (0.7)²
E = 0.000343 J
The kinetic energy Kair did the air molecules gain from the falling coffee filter is :
E = 0.000343 - 0.0441
= 0.0437 J
So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J
Learn more about kinetic energy here:
brainly.com/question/8101588
#SPJ4
4 is the difference sorry if i got it wrong :( :(
Complete Question
A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?
Answer:
The values is 
Explanation:
From the question we are told that
The diameter is 
The charge is 
The distance from the center is 
Generally the radius is mathematically represented as

=> 
=> 
Generally electric field is mathematically represented as
![E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7BQ%7D%7B%202%5Cepsilon_o%20%7D%20%5B1%20-%20%5Cfrac%7Bk%7D%7B%5Csqrt%7Br%5E2%20%2B%20%20k%5E2%20%7D%20%7D%20%5D)
substituting values
![E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B4.4%20%2A10%5E%7B-9%7D%7D%7B%202%2A%20%288.85%2A10%5E%7B-12%7D%29%20%7D%20%5B1%20-%20%5Cfrac%7B%281.00%20%2A10%5E%7B-4%7D%29%7D%7B%5Csqrt%7B%280.06%29%5E2%20%2B%20%20%281.0%2A10%5E%7B-4%7D%29%5E2%20%7D%20%7D%20%5D)
