A man goes 30 km due north and then 40 km due east. How far away is he from his initial position.
Answers
Answered by
2
Step-by-step explanation:
Northern vector = 30
Eastern vector = 40
So resultant will decide the answer
R^2 = A^2 +B^2 +2AB cos theta
but here theta = 90°
R^2 = A^2 + B^2
= 30^2 + 40^2
R^2 = 2500
R = 50
he is therefore 50 km far
Answered by
0
Answer:
Step-by-step explanation:
acc to math he is 70 km ffrom his position
but acc to physics ....
his displacement from his mesn position =30j+40i northeast
magnitude of his displacement=50km
40square + 30 square underoot
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