1. Vector B =(52.8)x + (-47.5)y. What is the magnitude of B?
2. Vector B=(52.8)x + (47.5)y. What is the direction of B in radians counter clockwise from the positive x axis?
1. sqrt [(52.8)^2 + (-47.5)^2] = 71.0
2. tan^-1 (47.5/52.8) = 42.0 degrees
To find the magnitude of a vector, you can use the formula for the magnitude of a vector in two dimensions. The magnitude of a vector is the square root of the sum of the squares of its components.
For the first question, Vector B is given as B = (52.8)x + (-47.5)y. To find the magnitude of B, you need to take the square root of the sum of the squares of the x and y components.
Magnitude of B = sqrt[(52.8)^2 + (-47.5)^2]
= sqrt[2,790.24 + 2,256.25]
= sqrt[5,046.49]
= 71.0
Therefore, the magnitude of Vector B is 71.0.
For the second question, Vector B is given as B = (52.8)x + (47.5)y. To find the direction of B in radians counter-clockwise from the positive x-axis, you can use the arctan function.
Direction of B = arctan(47.5/52.8)
To convert the result to radians, you need to multiply the result by pi/180, since there are 180 degrees in pi radians.
Direction of B (in radians) = (arctan(47.5/52.8)) * (pi/180)
≈ 0.7327 radians
Therefore, the direction of Vector B in radians counter-clockwise from the positive x-axis is approximately 0.7327 radians.