Scalar quantities zpqngdm pg1 - MEDIUM

Scalar quantities have only magnitude, while vector quantities have both magnitude and direction.

The total length of the path traveled, found by summing the distances of each segment ($5 + 3$ km in this case).

The overall change in position, considering direction (6 km East).

The difference between the final and initial volumes (3 $m^3$).

Scalar quantities can be subtracted directly using arithmetic.

The sum of the individual masses (17 kg).

Finding the difference between final and initial temperatures.

Adding up the individual lengths of each segment of a path.

The result of subtracting the smaller quantity from the larger quantity (12 units).

Directly summing the individual time intervals (30 + 15 + 45).

The direction from the starting to ending point.

The direction of the acceleration.

The absolute value of the difference between the two forces (20 N).

Using the Pythagorean theorem with the eastward and northward distances as sides.

Calculating $\sqrt{100^2 + 50^2 + 20^2}$ to find total displacement.

Finding the final position based on movements in multiple directions.

Break the Southeast displacement into East and South components.

Calculating $200 obreak \cdot obreak \cos(30°)$.

$\sqrt{(5 \text{ m/s})^2 + (3 \text{ m/s})^2}$

Using Pythagorean theorem to find $\sqrt{30^2 + 10^2}$.

The resultant force is 7 N to the right.

7 N upwards.

To resolve a single force into two forces acting at right angles to each other.

The resultant force is zero.

When resolving a force into perpendicular components.

Applying the Pythagorean theorem: $\sqrt{40^2 + 30^2}$.

$20 \cdot \cos(30°)$

2 N to the left

The magnitude of the net force is directly obtained by subtracting.

3500 N