Answer:
Explanation:
Given
altitude of the Plane 
When Airplane is
away
Distance is changing at the rate of 
From diagram we can write as

differentiate above equation w.r.t time

as altitude is not changing therefore 

at 
substitute the value we get 

Answer:

Explanation:
We have,
The surface temperature of the star is 60,000 K
It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

Here,
= wavelength

So, the wavelength of the star is
.
Air and water have a good day
The first thing to do is to define the origin of the coordinate system as the point at which the moped journey begins.
Then, you must write the position vector:
r = -3j + 4i + 3j
Rewriting
r = 4i
To go back to where you started, you must go
d = -4i
That is to say, must travel a distance of 4Km to the west.
Answer
West, 4km.