Technical Sciences Grade 12 NSC September 2025 P1 (Eastern Cape) Questions and Answers

A: \(2mv\) in the opposite direction.

\(\Delta p = m(-v) - mv = -2mv\), so the change is \(2mv\) in the direction opposite to the original motion.

C: Graph C

With no air resistance, mechanical energy is conserved, so the graph of \(ME\) against \(t\) is a horizontal straight line (graph C).

B: Liquid pressure increases with an increase in depth.

Pressure in a liquid is given by \(P = \rho g h\), so it grows with depth and the base of the wall must be stronger.

C: Diagram C

A concave lens is a diverging lens: parallel rays spread out after passing through it (diagram C).

B: refraction away from the normal.

The angle of incidence is smaller than the critical angle, so there is no total internal reflection. Light moving from a denser to a less dense medium bends away from the normal.

D: Volt.

Emf is measured in volts, the same unit as potential difference.

A: DC generator.

A split-ring commutator gives direct current, and a machine that turns mechanical energy into electrical energy is a generator.

Draw the toolbox as a dot or box with four labelled arrows, all starting at the box:

1. Weight \(F_g\) (or \(mg\)), pointing straight down.
2. Normal force \(F_N\), pointing straight up.
3. Ntombi's applied force \(F_{Ntombi} = 150\) N, pointing up and to the right at \(60^\circ\) to the horizontal (or its components \(F_x\) horizontally and \(F_y\) vertically).
4. Frictional force \(f\), pointing horizontally to the left, opposite to the horizontal pull.

Arrows must touch the body, be solid lines and carry labels. No extra forces.

Newton's First Law of Motion.

An object stays at rest or keeps moving at constant velocity unless a non-zero net (resultant) force acts on it.

The horizontal component of Ntombi's pull is balanced by the static friction, so the net force on the toolbox is zero. With no net force, it stays at rest.

Tension is a pulling force that acts along a string or rope.

Example: the \(150\ \text{N}\) force Ntombi applies through the rope, \(F_{Ntombi}\).

DECREASES.

The vertical component of her pull reduces the normal force, and friction depends on the normal force.

Draw the box with three horizontal arrows only:

1. The horizontal component of Ntombi's pull, \(F_x = 150\cos 60^\circ\), pointing in the direction of motion (right).
2. Mncedisi's push, \(F_{Mncedisi} = 80\) N, also pointing right.
3. Kinetic friction \(f_k\), pointing left, opposite to the motion.

No vertical forces are shown.

Take the direction of motion as positive.

The normal force is reduced by the vertical part of the pull:

\( \mu_k = 0{,}03 \) (0,027)

Momentum is the product of an object's mass and its velocity.

\( v_i = 33{,}33\ \text{m}\cdot\text{s}^{-1} \), east

\( p = 18\,337\ \text{kg}\cdot\text{m}\cdot\text{s}^{-1} \), west

An isolated system is one on which the net external force is zero.

Compare the total kinetic energy before and after the collision.

\( \sum E_{ki} \neq \sum E_{kf} \), so kinetic energy is not conserved.

The collision is inelastic.

They are inversely proportional: the longer the contact time, the smaller the net force, since \( F_{net} = \frac{\Delta p}{\Delta t} \).

EQUAL TO

Three points explain how an airbag reduces injuries:

1. The impulse (change in momentum) of the occupant stays the same.
2. The airbag increases the contact time during the crash.
3. A longer contact time gives a smaller force on the occupant, so the injuries are less severe.

Take the direction towards the tree as negative for the force (it opposes the motion).

\( \Delta t = 0{,}30\ \text{s} \)

In an isolated system, the total mechanical energy (the sum of gravitational potential energy and kinetic energy) stays constant.

\( ME = 43{,}12\ \text{J} \)

Mechanical energy is conserved:

\( v = 9{,}39\ \text{m}\cdot\text{s}^{-1} \)

\( 43{,}12\ \text{J} \)

Mechanical energy stays constant, and at the ground the potential energy is zero, so all the mechanical energy is kinetic.

The box moves at constant velocity, so the rope force equals the weight.

\( F = 1960\ \text{N} \)

\( W = 5880\ \text{J} \)

\( P = 7840\ \text{W} \)

Check: \( \Delta t = 3 \div 4 = 0{,}75 \) s and \( P = \frac{5880}{0{,}75} = 7840 \) W.

A deforming force is a force that changes the shape and/or size of a body.

The wire has diameter \(2{,}0\ \text{mm}\), so its radius is \(0{,}001\ \text{m}\).

\( \sigma = 3{,}12 \times 10^{7}\ \text{Pa} \)

\( \varepsilon = 1{,}67 \times 10^{-4} \) (no unit)

A perfectly elastic body regains its original shape and size when the deforming force is removed.

Example: a steel spring (or steel).

Refraction

Refraction is the bending of light when it passes from one medium into another.

Behind the object, on the same side of the lens as the object (between the lens and F).

The three characteristics of the image are:

1. It is virtual, so it cannot be projected on a screen.
2. It is upright, with the same orientation as the object.
3. It is magnified, so it is larger than the object.

Two applications of a convex lens are:

1. A telescope.
2. Binoculars (a magnifying glass or a camera lens is also correct).

At infinity (no image is formed).

The object sits exactly at the focal point, so the rays leave the lens parallel to each other. Parallel rays never meet, so no real image forms.

An electromagnetic wave is a changing electric field and magnetic field that are perpendicular to each other and to the direction in which the wave travels.

\( \lambda = 1{,}0 \times 10^{-8}\ \text{m} \)

Two uses of X-rays are:

1. Radiography, to see inside the body, for example to check for broken bones.
2. Radiotherapy, to treat cancer (use in airport and other security scanners is also correct).

Capacitance is the amount of charge a capacitor can store per volt of potential difference.

Farad

\( A = 2{,}26 \times 10^{5}\ \text{m}^2 \)

Three ways to increase the capacitance of a capacitor are:

1. Decrease the distance between the plates.
2. Increase the area of the plates.
3. Use a dielectric material with a higher dielectric constant.

The potential difference across a conductor is directly proportional to the current in it, at constant temperature.

The \(25\ \Omega\) and \(30\ \Omega\) resistors are in series: \(R_s = 25 + 30 = 55\ \Omega\). This is in parallel with the \(80\ \Omega\) resistor:

\( I = 0{,}23\ \text{A} \)

\( W = 126{,}96\ \text{J} \)

Time used in the month: \( \frac{7{,}5}{60} \times 30 = 3{,}75\ \text{h} \).

\( \text{Cost} = \text{R}18{,}00 \)

Faraday's law of electromagnetic induction.

When the magnetic flux linked with a coil changes, an emf is induced in the coil. The size of the induced emf is directly proportional to the rate of change of the magnetic flux.

\( \Delta\phi = -0{,}91\ \text{Wb} \) (a change of \(0{,}91\ \text{Wb}\) in magnitude)