The distance between city a and city b is 833.345 miles.
We know that
1°=60'
The distance of city a from the initial ray is calculated as
x_a=3960*tan45.46°=4024.101 miles
The distance of city b from the initial ray is calculated as
x_b=3960*tan 38.86°=3190.75 miles
Now the distance between city a and b is equal to
4024.101-3190.75=833.345 miles
This is the vertical distance between the cities.
The magnitude of the net displacement is 95.3 m
Explanation:
To find the magnitude of the net displacement, we have to resolve each of the two displacements into the horizontal and vertical direction first.
1st displacement is:
at 
So its components are

2nd displacement is:
at 
So its components are

Therefore, the x- and y-components of the net displacement are:

Therefore, the magnitude of the final displacement is:

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molecule The clues are that it is elements, so that means atoms, and that it is
in a fixed proportion like a molecular formula.
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Answer:
A. Vx = 3.63 m/s
B. Vy = -45.73 m/s
C. |V| = 45.87 m/s
D. θ = -85.46°
Explanation:
Given that position, r, is given as:
r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk
Velocity is the derivative of position, r:
V = dr/dt = 3.63 - 11.46t^j
A. x component of velocity, Vx = 3.63 m/s
B. y component of velocity, Vy = -11.46t
t = 3.99 secs,
Vy = - 11.46 * 3.99 = -45.73 m/s
C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]
|V| = √(2091.2329 + 13.1769)
|V| = √(2104.4098)
|V| = 45.87 m/s
D. Angle of the velocity relative to the x axis, θ is given as:
tanθ = Vy/Vx
tanθ = -45.73/3.63
tanθ = -12.6
θ = -85.46°