Assume that a particle moves in a velocity field defined by V (x, y) = (x + 6)i + (y − 3)j. Find the path of a particle with ini
1 answer:
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Answer:
12.97 km
Explanation:
In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.
Taking north as positive y direction and east as positive x-direction, we have:
- Displacement 1: 2.00 km to the north
So
- Displacement 2: 60.0° south of east for 7.00 km
So
- Displacement 3: 9.50 km 35.0° north of east
So
So the net displacement along the two directions is:
So, the distance between the initial and final position is equal to the magnitude of the net displacement:
The answer:
the full question is as follow:
<span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D?
As shown in the figure,
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
D = 0 - (A + B + C) ,
let W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics
Magnitude
A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
the direction of D can be found by using Dx and Dy value
we know that tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°
direction is 26.02°
the full question is as follow:
<span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D?
As shown in the figure,
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
D = 0 - (A + B + C) ,
let W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics
Magnitude
A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
the direction of D can be found by using Dx and Dy value
we know that tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°
direction is 26.02°