Vector A has magnitude of 15.0 m/s and is 75° counter-clockwise up from the x-axis. What are the x- and y-components of the vect
or?
1 answer:
Answer:
x component 3.88 y- component 14.488
Explanation:
We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure
Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )
y- component will be 15 sin 75°=14.488
For verification the resultant of x and y component should be equal to 15
So
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Answer:
The magnitude of the resultant displacement is 21 mi and its direction is 16.7° north of west
Explanation:
Hi there!
Please see the figure for a better understanding of the problem. The total displacement vector will be the sum of both displacements:
The vector for the first displacement is:
First displacement = (20 mi, 0)
The second displacement:
Second displacement = (0, 6.0 mi)
The resultant displacement will be:
R = (20 mi, 0) + (0, 6.0 mi) = (20 mi + 0, 0 + 6.0 mi) = (20 mi, 6.0 mi)
The magnitude of this vector will be:
The magnitude of the vector displacement is 21 mi.
To find the direction of the vector R, we have to apply trigonometry:
In a right triangle the following trigonometric rule applies:
cos θ = adjacent side to the angle/ hypotenuse
In this case:
cos θ = 20 mi / magnitude of R
θ = 16.7°
The direction of the vector is 16.7° north of west.
