A plane is flying to Minnesota with a velocity of 277.73 km/h, N. The plane
1 answer:
Answer:
466 km/h
255°
measured anticlockwise from the east direction
Explanation:
find the resultant velocity of a plane after a crosswind affects its motion. take the positive
x -direction as east, and the positive y -direction as north. The components of the plane's velocity (without wind) are
- <em>upx =</em> 0
- <em>upy = 450 Km/ h</em>
<em>And for the wind.</em>
- <em>uwx = - 120 Km/h</em>
- <em>uwy = 0</em>
<em>The components for the r</em><em>esultant velocity </em><em>of the plane are given by</em>
- <em>ux =upx + uwx = 0 - 120 Km/h = </em><u><em>- 120 Km/h</em></u>
- <em>uy = upy + uwy = - 450 Km/h +0 = </em><u><em>- 450 Km/h</em></u>
<em>The </em><em>magnitude </em><em>of the resultant velocity is</em>
<em />
= 466 km/h
Hope this helps u : )
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Explanation:
Given parameters:
Distance = 15miles north = 24140.2m
Initial velocity = 0m/s
Final velocity = 4m/s
Unknown:
Speed, velocity and acceleration = ?
Solution:
The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.
Speed =
The speed of the student is 4m/s
Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;
Velocity =
The velocity of the student is 4m/s due north
Acceleration is the change in velocity with time;
To find the acceleration, we use
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
4² = 0² + 2x a x 24140.2
a = = 0.00033m/s²