Vector A, a wind velocity vector, has a magnitude of 20 miles per hour and is blowing toward the North. A second wind velocity vector, Vector B, has a magnitude of 50 miles per hour and is blowing toward the Southeast. A third wind velocity vector, Vector C, has a magnitude of 10 miles per hour and is blowing toward the Southwest These three winds emerge (come out of) three separate canyons and merge (meet) at a flagpole carrying a cloth flag. What direction (angle with respect to zero degrees) will the flag be pointing as a result of the three winds? Also, what is the magnitude of the resulting wind velocity vector (expressed in m/s)? Note: Convert all variables to the metric (mks) system before performing your calculations.

https://www.jiskha.com/display.cgi?id=1507135091

Please let me know if its correct or not.

=20N+(-49.804N)+4.418E+3.673N+(-9.301)(-E)?

To find the direction and magnitude of the resulting wind velocity vector, we can first convert all the given velocities to the metric system (m/s).

Given:
- Vector A magnitude: 20 miles per hour
- Vector B magnitude: 50 miles per hour
- Vector C magnitude: 10 miles per hour

First, we need to convert miles per hour to meters per second. We know that 1 mile is equal to 1609.34 meters, and 1 hour is equal to 3600 seconds.

Converting Vector A:
20 mph * (1609.34 m/1 mile) * (1 hour/3600 s) ≈ 8.942 m/s

Converting Vector B:
50 mph * (1609.34 m/1 mile) * (1 hour/3600 s) ≈ 22.352 m/s

Converting Vector C:
10 mph * (1609.34 m/1 mile) * (1 hour/3600 s) ≈ 4.470 m/s

Now that we have the magnitudes of the three wind velocity vectors in the metric system, we can find the components of these vectors in the x and y directions.

Vector A: North Direction
- Magnitude (A): 8.942 m/s
- Direction (θ_A): 0 degrees (blowing toward the North)

Vector B: Southeast Direction
- Magnitude (B): 22.352 m/s
- Direction (θ_B): 45 degrees (blowing toward the Southeast)

Vector C: Southwest Direction
- Magnitude (C): 4.470 m/s
- Direction (θ_C): 225 degrees (blowing toward the Southwest)

Now, we can find the x and y components of these vectors.
For Vector A, since it is blowing towards the North, its x-component (A_x) is zero, and the y-component (A_y) is A (8.942 m/s).

For Vector B, since it is blowing toward the Southeast at a 45-degree angle, we can find its x-component (B_x) and y-component (B_y) using trigonometry.
B_x = B * cos(45 degrees) ≈ 22.352 m/s * cos(45 degrees) ≈ 15.822 m/s
B_y = B * sin(45 degrees) ≈ 22.352 m/s * sin(45 degrees) ≈ 15.822 m/s

Similarly, for Vector C, since it is blowing toward the Southwest at a 225-degree angle:
C_x = C * cos(225 degrees) ≈ 4.470 m/s * cos(225 degrees) ≈ -3.155 m/s (negative sign indicates direction opposite to the positive x-axis)
C_y = C * sin(225 degrees) ≈ 4.470 m/s * sin(225 degrees) ≈ -3.155 m/s (negative sign indicates direction opposite to the positive y-axis)

To find the resulting wind velocity vector, we need to sum up the x and y components of all three vectors.

Resulting x-component (R_x) = A_x + B_x + C_x = 0 + (15.822 m/s) + (-3.155 m/s) ≈ 12.667 m/s
Resulting y-component (R_y) = A_y + B_y + C_y = (8.942 m/s) + (15.822 m/s) + (-3.155 m/s) ≈ 21.609 m/s

Now, we can find the magnitude (R) and direction (θ_R) of the resulting wind velocity vector using the x and y components.

Magnitude of the resulting wind velocity vector (R) ≈ sqrt(R_x^2 + R_y^2) ≈ sqrt((12.667 m/s)^2 + (21.609 m/s)^2) ≈ 25.146 m/s

Direction (θ_R) = atan(R_y/R_x) ≈ atan(21.609 m/s / 12.667 m/s) ≈ 59.98 degrees

Therefore, the flag will be pointing at an angle of approximately 59.98 degrees with respect to zero degrees (positive x-axis). The magnitude of the resulting wind velocity vector is approximately 25.146 m/s.