**Explanation:** Velocity is a vector quantity because it has both magnitude and direction, unlike speed, which is a scalar quantity and only has magnitude. Distance and mass are also scalar quantities.

**Explanation:** This formula represents the position vector \(\vec{r}\) in two-dimensional motion, where \(\vec{r_0}\) is the initial position vector, \(\vec{v}\) is the initial velocity vector, \(\vec{a}\) is the acceleration vector, and \(t\) is the time.

**Explanation:** The trajectory of a projectile under the influence of gravity, without air resistance, is a parabola. This is because the horizontal and vertical motions are independent of each other, with the vertical motion being uniformly accelerated due to gravity.

**Explanation:** The horizontal component of velocity remains constant throughout the projectile’s motion because there are no horizontal forces acting on the projectile (assuming no air resistance).

**Explanation:** For maximum horizontal range, the angle of projection should be 45 degrees. This angle provides the optimal balance between the horizontal and vertical components of the initial velocity.

**Explanation:** In projectile motion, the horizontal component of velocity remains constant if air resistance is neglected because there are no horizontal forces acting on the object.

**Explanation:** The time of flight for a projectile launched at an angle θ with initial velocity \(v_0\) is given by \(T = \frac{2v_0 \sin\theta}{g}\), where \(g\) is the acceleration due to gravity. This formula considers the total time the projectile spends in the air.

**Explanation:** In projectile motion, the vertical acceleration remains constant and equal to \(g\), the acceleration due to gravity. The horizontal acceleration is zero in the absence of air resistance.

**Explanation:** In uniform circular motion, the centripetal acceleration always points toward the center of the circle, keeping the object in circular motion.

**Explanation:** The magnitude of the centripetal acceleration \(a_c\) for an object moving in a circle of radius \(r\) with a constant speed \(v\) is given by \(a_c = \frac{v^2}{r}\). This acceleration is directed toward the center of the circle.

**Explanation:** Scalar quantities are defined by their magnitude alone, such as distance and speed, while vector quantities have both magnitude and direction, such as displacement and velocity.

**Explanation:** Force is a vector quantity because it has both magnitude and direction. Time and temperature are scalar quantities with only magnitude, and distance is also a scalar quantity.

**Explanation:** A vector quantity is represented by a numerical value (magnitude), a direction, and a unit of measurement, for example, 5 meters per second north.

**Explanation:** In two-dimensional motion, the horizontal and vertical components of velocity are perpendicular to each other. They can be independently analyzed using vector addition.

**Explanation:** The graphical method for finding the resultant of two vectors involves placing the tail of the second vector at the head of the first vector and then drawing the resultant vector from the tail of the first vector to the head of the second vector.

**Explanation:** When two vectors of equal magnitude but opposite directions are added, their effects cancel each other out, resulting in a zero vector (no magnitude and no direction).

**Explanation:** The trajectory is the path that an object follows as it moves through space. In the context of motion in a plane, it is the two-dimensional path traced by the object.

**Explanation:** Relative velocity is the velocity of one object as observed from another moving object. It describes how fast one object is moving in relation to another.

**Explanation:** If an object moves with constant velocity in a plane, its trajectory is a straight line (linear). The object covers equal distances in equal time intervals in a specified direction.

**Explanation:** When a vector is multiplied by a scalar, the result is a vector whose magnitude is scaled by the scalar, but the direction remains the same as the original vector.

**Explanation:** Rectilinear motion is the motion of an object along a straight line. This type of motion is characterized by the object moving in a single dimension without changing its direction.

**Explanation:** A car moving on a straight road is an example of rectilinear motion, where the car moves in a straight line without deviating from its path.

**Explanation:** In rectilinear motion, average velocity \(v\) is calculated as the total displacement \(d\) divided by the total time \(t\) taken. This formula applies when the motion is uniform or the average is considered over a period.

**Explanation:** If a particle moves with uniform velocity, it means that there is no change in its velocity. Therefore, its acceleration, which is the rate of change of velocity, is zero.

**Explanation:** The equation \( s = ut + \frac{1}{2}at^2 \) represents the displacement \(s\) in uniformly accelerated rectilinear motion, where \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.

**Explanation:** Acceleration is the term used to describe the change in velocity per unit time. It is a vector quantity and can be positive (increasing velocity) or negative (decreasing velocity).

**Explanation:** The area under a velocity-time graph in rectilinear motion represents the displacement of the object. The graph visually depicts how the velocity changes over time, and the area calculates the total displacement.

**Explanation:** Distance is a scalar quantity that represents the total path traveled by an object, while displacement is a vector quantity that represents the shortest path between the initial and final positions of the object.

**Explanation:** If an object covers equal distances in equal intervals of time, it exhibits uniform motion. This indicates that the object’s speed is constant throughout its motion.

**Explanation:** Using the equation of motion \( v = u + at \), where \(u\) is the initial velocity (0 m/s), \(a\) is the acceleration (2 m/s²), and \(t\) is the time (5 seconds), we get \( v = 0 + (2 \times 5) = 10 \) m/s. So, the velocity after 5 seconds is 10 m/s.

**Explanation:** Curvilinear motion refers to the motion of an object along a curved path. It can be in any plane and is characterized by continuous change in direction.

**Explanation:** A stone thrown horizontally from a height follows a parabolic trajectory, which is an example of curvilinear motion due to the influence of gravity causing the path to curve.

**Explanation:** Tangential acceleration refers to the rate of change of the velocity along the path of the curve. It is responsible for changing the speed of the object as it moves along the curved path.

**Explanation:** Radial acceleration (also known as centripetal acceleration) is responsible for changing the direction of velocity in curvilinear motion. It acts towards the center of the curvature, causing the object to change its direction.

**Explanation:** The path of a projectile under the influence of gravity follows a parabolic trajectory. This is a result of the constant acceleration due to gravity acting downward while the horizontal component of motion remains constant.

**Explanation:** In uniform circular motion, the speed of the object remains constant, while its velocity changes continuously due to the change in direction. The centripetal acceleration keeps the object moving in a circular path.

**Explanation:** Centripetal force is required to maintain an object in circular motion. This force acts towards the center of the circle, causing the object to change direction continuously and move along a circular path.

**Explanation:** Non-uniform circular motion occurs when an object moves along a curved path with changing speed. Both the magnitude and direction of the velocity change in this type of motion.

**Explanation:** Angular velocity is the term used to describe the angle swept by the radius vector per unit time in circular motion. It measures how quickly an object rotates or revolves around a central point.

**Explanation:** The magnitude of centripetal acceleration \( a_c \) for an object moving with a constant speed \( v \) along a circular path with radius \( r \) is given by \( a_c = \frac{v^2}{r} \). This acceleration acts towards the center of the circle, keeping the object in circular motion.

**Explanation:** Studying motion in two dimensions allows us to understand and analyze more complex motions, such as projectile motion and circular motion, which cannot be adequately described by one-dimensional analysis alone.

**Explanation:** A ball thrown in the air follows a curved path and experiences both horizontal and vertical motion simultaneously, making it a clear example of two-dimensional motion.

**Explanation:** Studying motion in two dimensions helps us to separate and analyze the horizontal and vertical components of projectile motion independently, making it easier to predict the path of the projectile.

**Explanation:** Understanding motion in two dimensions is crucial for engineers when designing trajectories for projectiles, vehicles, and other objects, ensuring accurate predictions and safe, efficient designs.

**Explanation:** The study of two-dimensional motion is particularly important in astronomy for analyzing the orbits of planets, moons, and other celestial bodies, which often move in elliptical or curved paths.

**Explanation:** In sports science, understanding motion in two dimensions is vital for analyzing and optimizing the trajectories of balls in sports like basketball, soccer, and tennis, improving performance and strategy.

**Explanation:** A common method to analyze two-dimensional motion is vector decomposition, where the motion is broken down into perpendicular components (usually horizontal and vertical), making it easier to study and solve problems.

**Explanation:** A car turning around a curve on a flat road involves both forward motion and a change in direction, making it an example of two-dimensional motion.

**Explanation:** In aviation, studying motion in two dimensions is essential for plotting flight paths and understanding the dynamics of takeoff, landing, and maneuvering in the air.

**Explanation:** For satellite technology, understanding motion in two dimensions is crucial for calculating precise orbits and trajectories, ensuring that satellites are placed and maintained in their correct positions for optimal functionality.

**Explanation:** A scalar quantity is defined by its magnitude alone and does not include any information about direction. Examples include mass, temperature, and time.

**Explanation:** A vector quantity has both magnitude and direction. Velocity is a vector because it specifies the speed of an object as well as the direction of its motion.

**Explanation:** Scalar quantities can be added, subtracted, multiplied, and divided using standard arithmetic operations since they have only magnitude.

**Explanation:** A vector quantity is characterized by having both a magnitude and a direction, making it essential for representing quantities like force, velocity, and displacement.

**Explanation:** Work is a scalar quantity because it only involves magnitude (the amount of energy transferred) and does not depend on direction.

**Explanation:** Vectors are represented by arrows in diagrams, where the length of the arrow indicates the magnitude and the direction of the arrow indicates the direction of the vector.

**Explanation:** When two vectors are added together using the head-to-tail method, the resultant is another vector that represents the combined effect of the original vectors.

**Explanation:** The magnitude of a vector is represented by the length of the arrow in vector diagrams, with longer arrows indicating larger magnitudes.

**Explanation:** A unit vector has a magnitude of one and is used to indicate direction without considering magnitude. It is often used to normalize other vectors.

**Explanation:** The dot product (or scalar product) of two vectors is used to find the component of one vector along the direction of another. It results in a scalar quantity and is calculated as the product of the magnitudes of the vectors and the cosine of the angle between them.

**Explanation:** Temperature is a scalar quantity because it has only magnitude and no direction. It measures the degree of hotness or coldness of an object.

**Explanation:** The kilogram is the SI unit of mass. It measures the amount of matter in an object.

**Explanation:** The magnitude of a scalar quantity represents the size or amount of the quantity, such as the amount of mass or the degree of temperature.

**Explanation:** Acceleration is a vector quantity because it has both magnitude and direction, indicating the rate of change of velocity.

**Explanation:** Temperature is commonly measured in degrees Celsius (°C) or Fahrenheit (°F). The Kelvin (K) is the SI unit for temperature.

**Explanation:** A scalar quantity is defined solely by its magnitude and does not include any information about direction.

**Explanation:** Mass is a scalar quantity that remains constant regardless of location. It is not dependent on gravity, unlike weight.

**Explanation:** In scientific contexts, temperature is often expressed in Kelvin (K), which is the SI base unit for temperature and is used to measure absolute temperature.

**Explanation:** Temperature is a scalar quantity and does not change with direction. It only has magnitude, indicating how hot or cold something is.

**Explanation:** The mass of an object is the amount of matter it contains. Unlike weight, which is a force that depends on gravity, mass is a scalar quantity and does not change with location.

**Explanation:** Displacement is a vector quantity because it has both magnitude and direction. It represents the change in position of an object.

**Explanation:** Vector quantities have both magnitude and direction, unlike scalar quantities, which have only magnitude.

**Explanation:** Velocity is a vector quantity because it specifies both the speed of an object and the direction of its motion.

**Explanation:** Displacement is a vector quantity that represents the shortest path between two points and includes direction, while distance is a scalar quantity that measures the total path length traveled, regardless of direction.

**Explanation:** The SI unit of velocity is meter per second (m/s). Velocity includes both the speed of an object and its direction of motion.

**Explanation:** Velocity is defined as the speed of an object in a specific direction, making it a vector quantity.

**Explanation:** Two vectors can be added using the head-to-tail method or the parallelogram method, both of which take into account their magnitudes and directions.

**Explanation:** Division by a vector is not a standard operation in vector mathematics. Vectors can be added, subtracted, and multiplied by scalars, but division by a vector is not defined.

**Explanation:** When a vector is multiplied by a scalar, the resulting vector has the same direction as the original vector but its magnitude is scaled by the scalar.

**Explanation:** Velocity measures how fast an object is moving in a specific direction. Unlike speed, which is a scalar quantity, velocity is a vector quantity that includes direction.

**Explanation:** The magnitude of a vector is represented by the length of the arrow in a diagram. Longer arrows indicate larger magnitudes.

**Explanation:** The direction of a vector is indicated by the orientation of the arrow in the vector diagram.

**Explanation:** A vector arrow in a vector diagram conveys both the magnitude (length of the arrow) and the direction (orientation of the arrow).

**Explanation:** In a two-dimensional plane, the direction of a vector can be specified by its coordinates (x, y) or by the angle it makes with a reference axis, usually the x-axis.

**Explanation:** A unit vector is a vector with a magnitude of one. It is used to indicate direction without considering magnitude.

**Explanation:** A unit vector is typically denoted with a hat or caret symbol (^) above the variable, such as \(\hat{i}\), \(\hat{j}\), or \(\hat{k}\).

**Explanation:** The magnitude of vector \(\mathbf{A}\) is found using the Pythagorean theorem: \(\sqrt{A_x^2 + A_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).

**Explanation:** The angle of a vector in a vector diagram represents the direction of the vector relative to a reference axis, typically the x-axis.

**Explanation:** A vector with a magnitude of 10 units and an angle of 30 degrees from the positive x-axis is represented as \(10 \cos(30^\circ) \hat{i} + 10 \sin(30^\circ) \hat{j}\).

**Explanation:** In a three-dimensional space, vectors are represented algebraically by using three scalar components, typically denoted as \(x\), \(y\), and \(z\).

**Explanation:** The horizontal and vertical components of a vector are used to break down the vector into perpendicular directions, usually along the x and y axes in a coordinate system.

**Explanation:** The horizontal component of a vector \(\mathbf{A}\) is given by \(A \cos(\theta)\), where \(A\) is the magnitude of the vector and \(\theta\) is the angle with the horizontal axis.

**Explanation:** The vertical component of a vector \(\mathbf{A}\) is given by \(A \sin(\theta)\), where \(A\) is the magnitude of the vector and \(\theta\) is the angle with the horizontal axis.

**Explanation:** For a vector with magnitude 5 units and angle 60 degrees, the horizontal component is \(5 \cos(60^\circ) = 5 \times \frac{1}{2} = 2.5\) units and the vertical component is \(5 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} = 2.5\sqrt{3}\) units.

**Explanation:** The magnitude of the vector is found using the Pythagorean theorem: \(\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\) units.

**Explanation:** The angle \(\theta\) is given by \( \tan(\theta) = \frac{4}{3} \). Therefore, \(\theta = \tan^{-1}(\frac{4}{3})\).

**Explanation:** The horizontal component is given by \(B \cos(30^\circ)\). Therefore, \(B = \frac{6}{\cos(30^\circ)} = \frac{6}{\frac{\sqrt{3}}{2}} = 6 \times \frac{2}{\sqrt{3}} = 12\) units.

**Explanation:** The vertical component is given by \(C \sin(45^\circ)\). Therefore, \(C = \frac{8}{\sin(45^\circ)} = \frac{8}{\frac{\sqrt{2}}{2}} = 8 \times \frac{2}{\sqrt{2}} = 8\sqrt{2}\) units.

**Explanation:** The resultant vector can be found using the Pythagorean theorem, where the magnitude of the resultant vector is the square root of the sum of the squares of the horizontal and vertical components.

**Explanation:** For a vector with magnitude 10 units and angle 45 degrees, the horizontal and vertical components are \(10 \cos(45^\circ) = 10 \times \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}}\) units and \(10 \sin(45^\circ) = 10 \times \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}}\) units.

**Explanation:** Scalar multiplication of a vector by a scalar quantity increases the magnitude of the vector proportionally.

**Explanation:** Scalar multiplication \(2\mathbf{A}\) multiplies the magnitude of vector \(\mathbf{A}\) by 2. Therefore, if \(\mathbf{A}\) has a magnitude of 5 units, \(2\mathbf{A}\) will have a magnitude of \(2 \times 5 = 10\) units.

**Explanation:** Scalar multiplication \(-\frac{1}{2}\mathbf{B}\) multiplies the magnitude of vector \(\mathbf{B}\) by \(\frac{1}{2}\). Therefore, if \(\mathbf{B}\) has a magnitude of 10 units, \(-\frac{1}{2}\mathbf{B}\) will have a magnitude of \(\frac{1}{2} \times 10 = 5\) units.

**Explanation:** Scalar multiplication affects the magnitude of a vector but does not change its direction. The direction remains the same.

**Explanation:** Scalar multiplication \(2\mathbf{C}\) multiplies each component of vector \(\mathbf{C}\) by 2. Therefore, if \(C_x = 3\) and \(C_y = 4\), then \(2\mathbf{C}\) will have components \(2C_x = 6\) and \(2C_y = 8\).

**Explanation:** Scalar multiplication \(-\frac{1}{2}\mathbf{D}\) multiplies each component of vector \(\mathbf{D}\) by \(-\frac{1}{2}\). Therefore, if \(D_x = -2\) and \(D_y = 5\), then \(-\frac{1}{2}\mathbf{D}\) will have components \(-\frac{1}{2}D_x = -1\) and \(-\frac{1}{2}D_y = 2.5\).

**Explanation:** Scalar multiplication \(3\mathbf{E}\) multiplies each component of vector \(\mathbf{E}\) by 3. Therefore, if \(E_x = 4\) and \(E_y = 3\), then \(3\mathbf{E}\) will have components \(3E_x = 12\) and \(3E_y = 9\).

**Explanation:** Scalar multiplication \(\frac{1}{4}\mathbf{F}\) multiplies the magnitude of vector \(\mathbf{F}\) by \(\frac{1}{4}\). Therefore, if \(\mathbf{F}\) has a magnitude of 8 units, \(\frac{1}{4}\mathbf{F}\) will have a magnitude of \(\frac{1}{4} \times 8 = 2\) units.

**Explanation:** Scalar multiplication \(5\mathbf{G}\) multiplies each component of vector \(\mathbf{G}\) by 5. Therefore, if \(G_x = 6\) and \(G_y = 8\), then \(5\mathbf{G}\) will have components \(5G_x = 30\) and \(5G_y = 40\).

**Explanation:** Scalar multiplication of a vector by zero results in a vector with zero magnitude, regardless of its original direction.

**Explanation:** Graphically, when adding two vectors, the resultant vector is represented by the vector sum of the two original vectors.

**Explanation:** To add vectors graphically, the tail of the second vector is placed at the head of the first vector.

**Explanation:** When vectors are added graphically in the same direction, their lengths are added to find the magnitude of the resultant vector. Therefore, \(|\mathbf{A} + \mathbf{B}| = 5 \text{ cm} + 3 \text{ cm} = 8 \text{ cm}\).

**Explanation:** When vectors are added graphically in opposite directions, their lengths are subtracted to find the magnitude of the resultant vector. Therefore, \(|\mathbf{C} + \mathbf{D}| = |4 \text{ cm} – 6 \text{ cm}| = 2 \text{ cm}\).

**Explanation:** When vectors are added graphically at right angles (perpendicular vectors), the magnitude of the resultant vector is found using the Pythagorean theorem. Therefore, \(|\mathbf{E} + \mathbf{F}| = \sqrt{7^2 + 3^2} = \sqrt{49 + 9} = \sqrt{58} \approx 7.62 \text{ cm}\).

**Explanation:** When vectors are added graphically in opposite directions, their lengths are added to find the magnitude of the resultant vector. Therefore, \(|\mathbf{G} + \mathbf{H}| = |6 \text{ cm} + 8 \text{ cm}| = 14 \text{ cm}\).

**Explanation:** When vectors are added graphically in the same direction, their lengths are added to find the magnitude of the resultant vector. Therefore, \(|\mathbf{I} + \mathbf{J}| = 10 \text{ cm} + 5 \text{ cm} = 15 \text{ cm}\).

**Explanation:** When vectors are added graphically at an angle, the magnitude of the resultant vector can be found using the cosine rule or by resolving the vectors into components. Given that they are at an angle of 60 degrees, the magnitude of the resultant vector is \(|\mathbf{K} + \mathbf{L}| = \sqrt{12^2 + 9^2 + 2 \times 12 \times 9 \times \cos(60^\circ)} = \sqrt{144 + 81 + 108} = \sqrt{333} \approx 18.26 \text{ cm}\).

**Explanation:** When vectors are added graphically at an angle, the magnitude of the resultant vector can be found using the cosine rule or by resolving the vectors into components. Given that they are at an angle of 120 degrees, the magnitude of the resultant vector is \(|\mathbf{M} + \mathbf{N}| = \sqrt{8^2 + 6^2 + 2 \times 8 \times 6 \times \cos(120^\circ)} = \sqrt{64 + 36 – 48} = \sqrt{52} \approx 7.21 \text{ cm}\).

**Explanation:** When vectors are added graphically at right angles (perpendicular vectors), the magnitude of the resultant vector is found using the Pythagorean theorem. Therefore, \(|\mathbf{P} + \mathbf{Q}| = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} \approx 11.18 \text{ cm}\).

**Explanation:** The Triangle Law of Vector Addition states that to find the resultant of two vectors, you add them together vectorially.

**Explanation:** When vectors act along the same straight line, their resultant using the Triangle Law of Vector Addition is simply the sum of their magnitudes. Therefore, \(|\mathbf{A} + \mathbf{B}| = 5 \text{ units} + 3 \text{ units} = 8 \text{ units}\).

**Explanation:** When vectors act in opposite directions along the same straight line, their resultant using the Triangle Law of Vector Addition is the absolute difference of their magnitudes. Therefore, \(|\mathbf{C} + \mathbf{D}| = |4 \text{ units} – 6 \text{ units}| = 2 \text{ units}\).

**Explanation:** When vectors act at right angles to each other, the magnitude of their resultant using the Triangle Law of Vector Addition is found using the Pythagorean theorem. Therefore, \(|\mathbf{E} + \mathbf{F}| = \sqrt{7^2 + 3^2} = \sqrt{49 + 9} = \sqrt{58} \approx 7.62 \text{ units}\).

**Explanation:** When vectors act in opposite directions along the same straight line, their resultant using the Triangle Law of Vector Addition is the absolute sum of their magnitudes. Therefore, \(|\mathbf{G} + \mathbf{H}| = |6 \text{ units} + 8 \text{ units}| = 14 \text{ units}\).

**Explanation:** When vectors act at an angle to each other, the magnitude of their resultant using the Triangle Law of Vector Addition can be found using the cosine rule or by resolving the vectors into components. Given that they are at an angle of 120 degrees, \(|\mathbf{I} + \mathbf{J}| = \sqrt{10^2 + 5^2 + 2 \times 10 \times 5 \times \cos(120^\circ)} = \sqrt{100 + 25 – 50} = \sqrt{75} = 5\sqrt{3} \approx 8.66 \text{ units}\).

**Explanation:** When vectors act at an angle to each other, the magnitude of their resultant using the Triangle Law of Vector Addition can be found using the cosine rule or by resolving the vectors into components. Given that they are at an angle of 60 degrees, \(|\mathbf{K} + \mathbf{L}| = \sqrt{12^2 + 9^2 + 2 \times 12 \times 9 \times \cos(60^\circ)} = \sqrt{144 + 81 + 108} = \sqrt{333} \approx 18.26 \text{ units}\).

**Explanation:** When vectors act at an angle to each other, the magnitude of their resultant using the Triangle Law of Vector Addition can be found using the cosine rule or by resolving the vectors into components. Given that they are at an angle of 120 degrees, \(|\mathbf{M} + \mathbf{N}| = \sqrt{8^2 + 6^2 + 2 \times 8 \times 6 \times \cos(120^\circ)} = \sqrt{64 + 36 – 48} = \sqrt{52} \approx 7.21 \text{ units}\).

**Explanation:** When vectors act at right angles (perpendicular vectors), the magnitude of their resultant using the Triangle Law of Vector Addition can be found using the Pythagorean theorem. Therefore, \(|\mathbf{P} + \mathbf{Q}| = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} \approx 11.18 \text{ units}\).

**Explanation:** The horizontal component of \(\mathbf{V}\) can be found using \( V_x = V \cos(\theta) \), where \( V = 10 \) units and \( \theta = 30^\circ \). Therefore, \( V_x = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5 \) units. The vertical component \( V_y = V \sin(\theta) \) gives \( V_y = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \) units.

**Explanation:** The horizontal component of the force can be found using \( F_x = F \cos(\theta) \), where \( F = 20 \) N and \( \theta = 60^\circ \). Therefore, \( F_x = 20 \cos(60^\circ) = 20 \times \frac{1}{2} = 10 \) N. The vertical component \( F_y = F \sin(\theta) \) gives \( F_y = 20 \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \) N.

**Explanation:** The magnitude of vector \(\mathbf{V}\) can be found using \( |\mathbf{V}| = \sqrt{V_x^2 + V_y^2} \), where \( V_x = 8 \) m/s and \( V_y = 6 \) m/s. Therefore, \( |\mathbf{V}| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \) m/s.

**Explanation:** To find the vertical component \( D_y \) of vector \(\mathbf{D}\), use \( D_y = \sqrt{D^2 – D_x^2} \), where \( D = 12 \) units and \( D_x = 9 \) units. Therefore, \( D_y = \sqrt{12^2 – 9^2} = \sqrt{144 – 81} = \sqrt{63} \approx 8 \) units.

**Explanation:** The horizontal component of velocity can be found using \( V_x = V \cos(\theta) \), where \( V = 10 \) m/s and \( \theta = 45^\circ \). Therefore, \( V_x = 10 \cos(45^\circ) = 10 \times \frac{\sqrt{2}}{2} = 5\sqrt{2} \) m/s. The vertical component \( V_y = V \sin(\theta) \) gives \( V_y = 10 \sin(45^\circ) = 10 \times \frac{\sqrt{2}}{2} = 5\sqrt{2} \) m/s.

**Explanation:** The magnitude of vector \(\mathbf{F}\) can be found using \( |\mathbf{F}| = \sqrt{F_x^2 + F_y^2} \), where \( F_x = 15 \) N and \( F_y = 20 \) N. Therefore, \( |\mathbf{F}| = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \) N.

**Explanation:** To find the resultant vector \(\mathbf{A} + \mathbf{B}\), add the corresponding components: \( \mathbf{A} + \mathbf{B} = (3 + 2)\hat{i} + (-4 + 5)\hat{j} = 5\hat{i} + \hat{j} \).

**Explanation:** The magnitude of vector \(\mathbf{V}\) can be found using \( |\mathbf{V}| = \sqrt{V_x^2 + V_y^2} \), where \( V_x = 8 \) m and \( V_y = -6 \) m. Therefore, \( |\mathbf{V}| = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \) m.

**Explanation:** The x-component \(\mathbf{A}_x\) can be found using \( \mathbf{A}_x = A \cos(\theta) \) and the y-component \(\mathbf{A}_y\) using \( \mathbf{A}_y = A \sin(\theta) \), where \( A = 5 \) units and \( \theta = 60^\circ \). Therefore, \( \mathbf{A}_x = 5 \cos(60^\circ) = 5 \times \frac{1}{2} = 2.5 \) units and \( \mathbf{A}_y = 5 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} \approx 4.33 \) units.

**Explanation:** The angle \(\theta\) can be found using \( \theta = \tan^{-1}\left(\frac{\mathbf{B}_y}{\mathbf{B}_x}\right) \), where \( \mathbf{B}_x = 10 \) m and \( \mathbf{B}_y = -8 \) m. Therefore, \( \theta = \tan^{-1}\left(\frac{-8}{10}\right) = \tan^{-1}(-0.8) \approx -36.87^\circ \), but since angle is measured with the positive x-axis, so answer is positive 36.

**Explanation:** To find the resultant vector \(\mathbf{A} + \mathbf{B}\), add the corresponding components: \( \mathbf{A} + \mathbf{B} = (4 + 2)\hat{i} + (-3 + 5)\hat{j} = 6\hat{i} + 2\hat{j} \).

**Explanation:** To find the resultant force, add the corresponding components: \( \mathbf{F}_{\text{resultant}} = (10 – 3) \hat{i} + (-5 + 7) \hat{j} = 7 \hat{i} + 2 \hat{j} \).

**Explanation:** To find the resultant vector \(\mathbf{A} + \mathbf{B}\), add the corresponding components: \( \mathbf{A} + \mathbf{B} = (8 + 5) \hat{i} + (-6 + 4) \hat{j} = 13 \hat{i} – 2 \hat{j} \).

**Explanation:** To find the resultant velocity, add the corresponding components: \( \mathbf{V}_{\text{resultant}} = (5 – 2) \hat{i} + (3 + 4) \hat{j} = 3 \hat{i} – 1 \hat{j} \).

**Explanation:** To find the magnitude of the resultant vector, first find the components \(\mathbf{A} + \mathbf{B}\): \( \mathbf{A} + \mathbf{B} = (3 – 1) \hat{i} + (2 + 5) \hat{j} = 2 \hat{i} + 7 \hat{j} \). Then, \( |\mathbf{A} + \mathbf{B}| = \sqrt{(2)^2 + (7)^2} = \sqrt{4 + 49} = \sqrt{53} \).

**Explanation:** To find the resultant vector \(\mathbf{A} + \mathbf{B}\), add the corresponding components: \( \mathbf{A} + \mathbf{B} = (5 – 2) \hat{i} + (3 + 4) \hat{j} = 3 \hat{i} + 7 \hat{j} \).

**Explanation:** To find the resultant displacement, add the corresponding components: \( \mathbf{D}_{\text{resultant}} = (8 – 4) \hat{i} + (-3 + 6) \hat{j} = 4 \hat{i} + 3 \hat{j} \).

**Explanation:** To find the magnitude of the resultant vector, first find the components \(\mathbf{P} + \mathbf{Q}\): \( \mathbf{P} + \mathbf{Q} = (-3 + 7) \hat{i} + (5 – 2) \hat{j} = 4 \hat{i} + 3 \hat{j} \). Then, \( |\mathbf{P} + \mathbf{Q}| = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.

**Explanation:** To find the direction of the resultant velocity, use \( \theta = \tan^{-1}\left(\frac{V_{\text{resultant, } y}}{V_{\text{resultant, } x}}\right) \), where \( V_{\text{resultant, } x} = 6 – 3 = 3 \) m/s and \( V_{\text{resultant, } y} = -2 + 4 = 2 \) m/s. Therefore, \( \theta = \tan^{-1}\left(\frac{2}{3}\right) \).

**Explanation:** To find the resultant vector, add the corresponding components: \( \mathbf{U}_1 + \mathbf{U}_2 = 10 \hat{i} + 5 \hat{j} \).

**Explanation:** To find the resultant displacement, use the Pythagorean theorem: \( \text{Resultant displacement} = \sqrt{(10)^2 + (8)^2} = \sqrt{100 + 64} = \sqrt{164} \approx 12 \) km.

**Explanation:** To find the magnitude of the resultant force, use \( |\mathbf{F}_{\text{resultant}}| = \sqrt{(10 – 3)^2 + (-5 + 7)^2} = \sqrt{7^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53} \approx 11.2 \) N.

**Explanation:** Horizontal component: \(6 \cos 30^\circ = 6 \cdot \frac{\sqrt{3}}{2} \approx 5.2\) m/s, Vertical component: \(6 \sin 30^\circ = 6 \cdot \frac{1}{2} = 3\) m/s.

**Explanation:** Horizontal component: \(20 \cos 45^\circ = 20 \cdot \frac{\sqrt{2}}{2} = 10 \sqrt{2}\) m/s, Vertical component: \(20 \sin 45^\circ = 20 \cdot \frac{\sqrt{2}}{2} = 10 \sqrt{2}\) m/s.

**Explanation:** Horizontal component: \(250 \cos 60^\circ = 250 \cdot \frac{1}{2} = 125\) km/h, Vertical component: \(250 \sin 60^\circ = 250 \cdot \frac{\sqrt{3}}{2} \approx 125 \sqrt{3}\) km/h (south direction, so negative sign).

**Explanation:** Time of flight \(T = \frac{2V_0 \sin \theta}{g} = \frac{2 \cdot 20 \cdot \sin 30^\circ}{9.8} = \frac{40 \cdot 0.5}{9.8} = \frac{20}{9.8} \approx 2.04\) s. Rounded to nearest whole number, \(T \approx 4\) s.

**Explanation:** Horizontal range \(R = V_0 \cdot T = 15 \cdot \frac{2h}{g} = 15 \cdot \frac{2 \cdot 15}{10} = 30\) m.

**Explanation:** Centripetal acceleration \(a_c = \frac{v^2}{r} = \frac{10^2}{50} = \frac{100}{50} = 2\) m/s².

**Explanation:** Centripetal acceleration \(a_c = \frac{v^2}{r} = \frac{5^2}{4} = \frac{25}{4} = 6.25\) m/s².

**Explanation:** Centripetal acceleration \(a_c = \frac{v^2}{r} = \frac{v^2}{g} = \frac{10}{8} = 1.25\). Radius \(r = \frac{v^2}{a_c} = \frac{10}{1.25} = 8\) m

**Explanation:** In uniform circular motion, an object moves in a circle with a constant speed \(v\) and its velocity changes due to the direction of motion, always tangential to the circle.

**Explanation:** The acceleration \(a\) in uniform circular motion is given by \( a = \frac{v^2}{r} \), where \(v\) is the speed of the particle and \(r\) is the radius of the circle.

**Explanation:** The centripetal force required for uniform circular motion can be provided by various forces like tension, friction, or gravity, depending on the specific scenario.

**Explanation:** In uniform circular motion, the speed of the car remains constant, but its velocity changes continuously due to the change in direction.

**Explanation:** Centripetal force acts towards the center of the circular path and is responsible for keeping the object moving in a circle at a constant speed.

**Explanation:** Centripetal acceleration \(a_c = \frac{v^2}{r}\) points towards the center of the circle and depends on the square of the speed divided by the radius of the circle.

**Explanation:** Centripetal acceleration \(a_c = \frac{v^2}{r} = \frac{5^2}{10} = \frac{25}{10} = 2.5\) m/s².

**Explanation:** Centripetal force required for uniform circular motion can be provided by various forces like tension, friction, or gravity, depending on the specific scenario.

**Explanation:** Centripetal force \(F_c = m \cdot \frac{v^2}{r} = m \cdot \frac{20^2}{100} = m \cdot 4\) N, where \(m\) is the mass of the car.

**Explanation:** Centripetal force \(F_c = m \cdot \frac{v^2}{r}\). If \(v\) is doubled, \(v^2\) becomes four times larger, hence the centripetal force quadruples.

**Explanation:** Period \(T\) is the time taken for one complete cycle or revolution in circular motion.

**Explanation:** Period \(T = \frac{\text{Total time}}{\text{Number of revolutions}} = \frac{10}{50} = 0.2\) s per revolution.

**Explanation:** Frequency \(f\) is the number of cycles or revolutions per unit time, where \(T\) is the period.

**Explanation:** Frequency \(f = \frac{1}{T} = \frac{1}{2} = 0.5\) Hz, where \(T\) is the period in seconds.

**Explanation:** Angular velocity \(\omega = \frac{\theta}{t}\) represents the rate of change of angular displacement with respect to time.

**Explanation:** not available

**Explanation:** Angular velocity \(\omega\) is calculated using the formula \(\omega = \frac{2\pi n}{T}\), where \(n\) is the number of revolutions and \(T\) is the time taken. Here, \(\omega = \frac{2\pi \cdot 10}{5} = 4\) rad/s.

**Explanation:** Angular velocity is measured in radians per second (rad/s) in the SI unit system.

**Explanation:** Linear speed \(v\) is given by \(v = \omega \cdot r = \frac{\pi}{3} \cdot 4 = 4\pi\) m/s, where \(r\) is the radius and \(\omega\) is the angular velocity.

**Explanation:** Linear speed \(v = \omega \cdot r = 10 \cdot 0.5 = 5\) m/s. The linear speed is directly proportional to the radius and angular velocity of the rotating object.