Answer:
I believe the answer is speed up.
Explanation:
this is because when water heats up the molecules move father apart from each other they speed up, eventually causing the water to boll
The phenomenon which is responsible for this effect is called diffraction.
Diffraction is the ability of a wave to propagate when it meets an obstacle or a slit. When the wave encounters the obstacle or the slit, it 'bends' around it and it continues propagate beyond it. A classical example of this phenomenon is when a sound wave propagates through a wall where there is a small aperture (as in the example of this problem)
Answer:

Explanation:
Given:
- file size to be transmitted,

- transmission rate of data,

- propagation speed,

- distance of data transfer,

<u>Now the delay in data transfer from source to destination for each 10 Mb:</u>



<u>Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:</u>
So,



So, the average speed of the Cheetah is 17.6 m/s.
<h3>Introduction</h3>
Hello ! I'm Deva from Brainly Indonesia. This time, I will help regarding the average speed. The average speed is obtained from finding the average of the speeds that occur or can be detected from the division between distance and travel time. The average speed can be formulated by :

With the following condition :
= average speed (m/s)- s = shift or distance objects from initial movement (m)
- t = interval of the time (s)
<h3>Problem Solving</h3>
We know that :
- s = shift = 88 m
- t = interval of the time = 5 seconds
What was asked :
= average speed = ... m/s
Step by step :



So, the average speed of the Cheetah is 17.6 m/s.