A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 140 cm and makes an angle of 130° with the positive x axis. The resultant displacement has a magnitude of 144 cm and is directed at an angle of 30.0° to the positive x axis. Find the magnitude and direction of the second displacement.

To solve this problem, we can use vector addition.

Let's assume the first displacement is represented by vector A and the second displacement is represented by vector B. The resultant displacement is represented by vector R.

The given information tells us that the magnitude of vector A is 140 cm and it makes an angle of 130° with the positive x-axis.

To find the components of vector A, we can use trigonometry. The x-component of vector A is given by A_x = A * cosθ, where θ is the angle it makes with the positive x-axis. The y-component of vector A is given by A_y = A * sinθ.

Plugging in the values, we get:
A_x = 140 cm * cos(130°) ≈ -62.74 cm
A_y = 140 cm * sin(130°) ≈ 118.12 cm

Similarly, we can do the same for vector R. The magnitude of vector R is given as 144 cm and it makes an angle of 30.0° with the positive x-axis.

Using the same trigonometric formulas, we can find the x and y components of vector R:
R_x = 144 cm * cos(30.0°) = 124.6 cm
R_y = 144 cm * sin(30.0°) = 72.0 cm

Now, we can calculate the x and y components of vector B. Since vector R is the sum of vectors A and B, we can write the equation R = A + B. Breaking it down into x and y components, we get:
R_x = A_x + B_x => B_x = R_x - A_x
R_y = A_y + B_y => B_y = R_y - A_y

Plugging in the values, we have:
B_x = 124.6 cm - (-62.74 cm) ≈ 187.34 cm
B_y = 72.0 cm - 118.12 cm ≈ -46.12 cm

Now, we can find the magnitude and direction of vector B. The magnitude of vector B is given by its x and y components using Pythagoras' theorem:
|B| = √(B_x^2 + B_y^2)

Plugging in the values, we have:
|B| = √(187.34^2 + (-46.12)^2) ≈ 193.82 cm

To find the direction of vector B, we can use trigonometry again. The angle θ that vector B makes with the positive x-axis is given by:
θ = tan^(-1)(B_y / B_x)

Substituting the values, we have:
θ = tan^(-1)(-46.12 cm / 187.34 cm) ≈ -13.9°

Therefore, the magnitude of the second displacement is approximately 193.82 cm, and it makes an angle of approximately -13.9° with the positive x-axis. Note that the negative angle indicates that the direction is in the clockwise direction from the positive x-axis.