A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North.
Answer:
The troposphere starts at the Earth's surface and extends 8 to 14.5 kilometers high (5 to 9 miles). This part of the atmosphere is the most dense. Almost all weather is in this region.
Explanation:
To solve this problem, use the ratio given by the total number of electrons or protons that exist as a function of the total charge, and inversely proportional to the value of the fundamental charge. The number of fundamental unit 
that constitutes a charge of 40.0C can be calculated as
Here,
= Value of charge and it is the fundamental charge
Q = Total Charge
N = Total number of electron or protons
The number of fundamental units is calculated as follows
Therefore the number of fundamental charge units moved by lightning bolt is 
Distance travelled in south direction= 1.5hr*0.75km/hr= 1.125km
Distance travlled in north direction= 0.90*2.5=2.25
Net displacement = 2.25-1.125= 1.125 to the north
