12PHY Revision warm up

1. A student moves a magnet into a coil of wire as shown in the diagram. The current in the coil will create a magnetic field with North pointing

• Left

• Right

• Clockwise

• Anticlockwise

A spring oscillates freely up and down. At the bottom of it's oscillation, it has zero gravitational energy. At the top, it has zero spring energy. In the midpoint,

• Most of the energy is Kinetic. Spring and Kinetic share the rest.

• About half the energy is Gravitational. Kinetic and Spring share the rest.

• About half the energy is Kinetic. Spring and Gravitational share the rest.

• About half the energy is Spring Energy. Gravitational and Kinetic share the rest.

An armature is rotated in a B field as shown. At the moment shown in full colour, what direction will current flow through the ammeter?

• No Current

• Left to Right

• Right to Left

• Out of the page.

The connection type and output current are, respectively:

• Slip Rings, DC

• Split Ring Commutator, DC

• Slip Rings, AC

• Split Ring Commutator, AC

A rocket leaves Earth and travels to a star that is 20 light years away as measured by observatories. The speed of the rocket is 0.8c relative to the Earth. Through the journey, the astronauts will age by...

• 15 years

• 20 years

• 25 years

• 41.7 years

Which of the following experiments provides the clearest evidence of the wave nature of light?

• Fuzzy Shadows

• The Photoelectric Effect

• Light refraction

• Young's double-slit experiment

Which of the following experiments provides the clearest evidence of the particle nature of light?

• Fuzzy Shadows

• The Photoelectric Effect

• Light refraction

• Young's double-slit experiment

A car of mass 800 kg is towed along a straight road so that its velocity changes uniformly from 10 m s^-1 to 20 m s^-1 in a distance of 200 m. The frictional force is constant at 500 N.

Calculate the acceleration of the car.

_{VCAA 1977}

• 0.75 ms^-2

• 1.0 ms^-2

• 1.25 ms^-2

• 25 ms^-2

Answer explanation

v² = u² + 2ax

20² = 10² + 2 × a × 200

400 = 100 + 400a

300 = 400a

a = 0.75 ms^-2

Two blocks, each of mass m, are connected by means of a string which passes over a frictionless pulley. One is at rest on a frictionless table; the other is held at rest in the position shown.

What is the force of the string on Block X?

_{VCAA 1981}

• Zero

• mg/2

• mg

• 2mg

Answer explanation

Block X is initially stationary and remains stationary until the hand supporting Block Y is removed. Therefore the net force on Block X is zero

Two blocks, each of mass m, are connected by means of a string which passes over a frictionless pulley. One is at rest on a frictionless table; the other is held at rest in the position shown.

Block Y is then released, and the blocks move. What will then be the net force on Y?

_{VCAA 1981}

• Zero

• mg/2

• mg

• 2mg

Answer explanation

When Block Y is released, its weight accelerates it downward.

As the two blocks are connected by a string, they are going to both have the same acceleration when released.

The net force acting on Block Y is given by

mg – T = ma

The net force acting on Block X is given by

T = ma

Substituting for T gives

mg – ma = ma

                                   

mg = 2ma

                                   

ma = mg/2