Answer:
D. The disk in a computer hard drive
<em>The</em><em> </em><em>disk</em><em> </em><em>has</em><em> </em><em>a</em><em> </em><em>magnetic</em><em> </em><em>coating</em><em> </em><em>in</em><em> </em><em>which</em><em> </em><em>the</em><em> </em><em>infor</em><em>mation</em><em> </em><em>is</em><em> </em><em>stored </em><em>on</em><em> </em><em>the</em><em> </em><em>magnetic</em><em> </em><em>strips</em><em>.</em>
Answer:
A motorcycle changing speed from 20km/h to 35km/h
Explanation:
it doesn't show unbalanced force acting on the object but instead, the motorcycle changing speed from 20km/h to 35km/h
Answer:
b. slow-moving streams.
Explanation:
In Fluid Mechanics, the Reynolds numbers indicates the existence of turbulence in fluid streams. Low Reynolds numbers are related with laminar flow. The Reynolds formula is:

The Reynolds number is directly proportional to fluid speed. Hence, slow-moving streams are a sound example of laminar flow. The correct answer is B.
Answer:
The amplitude is
Explanation:
From the question we are told that
The frequency of when sound is approaching observer is 
The frequency as the move away from observer is 
The time between the pitch are 
Here you are the observer and your friends are the source of the sound
The period is mathematically evaluated as

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch
Now T can also be mathematically represented as

Where
is the angular velocity
=> 
=> 
Now using Doppler Effect,
The source of the sound is approaching the observer
The


Where A is the amplitude
So when the source is moving away from the observer
Here
is the fundamental frequency
Dividing the both equation we have




=> 

Answer:
answer is 2 option because more force is applied