Answer:
$ 3085713685.71
Explanation:
= Actual wavelength = 700 nm
= Changed wavelength = 500 nm
Let the wavelength of red color be 700 nm and green be 500 nm
Change in wavelength is

We have the relation

The speed of the vehicle is 85714285.7143 m/s

By how much was the car speeding

The number of 10 km/h in the above speed

Cost of the ticket

The cost of the ticket is $ 3085713685.71
If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled; and so on. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.
Explanation:
It is known that wave intensity is the power to area ratio.
Mathematically, I = 
As it is given that power is 28.0 W and area is
.
Therefore, sound intensity will be calculated as follows.
I = 
= 
= 
or, = 
Thus, we can conclude that sound intensity at the position of the microphone is
.