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




=> 

Because the airplane flies at 800 km/h, it will fly 800 km in 1 hour. In order to find how far it travels in 2 hours, we multiply this number by 2. 800*2 = 1600, so the plane will travel 1600 km in 2 hours.
A visual display of data or information is called a graph. There are many types of graphs. These can include pie graphs, bar graphs, and many more. Graphs are useful, because they show you visually data which is helpful to many. Hope this helped
Answer:
distance between object and image = 18.9 cm
Explanation:
given data
radius of curvature = 18 cm
focal length = 1/2 radius of curvature
magnification = 40%
to find out
distance between object and image
solution
we know lens formula that is
1/f = 1/v + 1/u ....................1
here f = 18 /2 and v and u is object and image distance
and we know m = 40% = 0.40
so 0.40 = -v / u
so here v = - 0.40 u
so from equation 1
1/f = 1/v + 1/u
2/18 = - 1/0.40u + 1/u
u = -13.5 cm ..................2
and
v = -0.40 (- 13.5)
v = 5.4 cm ......................3
so from equation 2 and 3
distance between object and image = 5.4 + 13.5
distance between object and image = 18.9 cm
During the winter, the Northern Hemisphere tilted away from the sun, receiving solar radiation at more of an angle. <u>This results in colder temperatures and more extreme temperature changes.</u>