Answer:
245.45km in a direction 21.45° west of north from city A
Explanation:
Let's place the origin of a coordinate system at city A.
The final position of the airplane is given by:
rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:
rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km
rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km
The module of this position is:

And the angle measure from the y-axis is:

So the answer is 245.45km in a direction 21.45° west of north from city A
Answer:
(A) Reading will be 65 N
(B) Net force on the elevator will be 49.076 N
Explanation:
We have given the balance force = 65 N
Acceleration due to gravity 
We know that W=mg
So 
m = 6.632 kg
(a) In first case as the as the speed is constant so the force on the elevator will be 65 N
(B) In second case as the elevator is decelerating at a rate of 
So net acceleration = 9.8-2.4=
So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N
Answer:
The answer
Explanation:
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Answer:
A. ls sensitive to heat
this is correct answer
I hope it's helpful for you.......
Answer:
KE = 1/2mv^2
KE = 1/2(24)(3^2)
KE = 12(9)
KE = 108 J
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