OCR A-Level Physics A Newtonian world and astrophysics overview quiz quiz

The energy supplied breaks intermolecular bonds rather than increasing kinetic energy

The substance simply stops absorbing any energy at all once it has reached its melting point

The specific heat capacity of the substance momentarily falls to zero while it is melting

The mass of the substance steadily decreases as it melts, so its temperature cannot rise

$pV = nRT$

$pV = NkT$

$pV = \frac{1}{3}Nm\langle c^2\rangle$

$pV = mc\Delta T$

It halves

It stays the same

It doubles

It quadruples

It is not accelerating at all, because its speed around the circle stays perfectly constant

The centripetal force does work on it, steadily increasing the kinetic energy of the object

The centripetal force points outwards, away from the centre, in the same sense as the motion

It accelerates towards the centre because its velocity direction changes

Acceleration is proportional to displacement and directed towards equilibrium

Acceleration is constant

Acceleration is proportional to displacement and directed away from equilibrium

Velocity is proportional to displacement

It decreases

It increases

It stays the same

It depends on the mass of the bob

It halves, since field strength is inversely proportional to distance

It doubles, since field strength is directly proportional to distance

It falls to one quarter

It is unchanged, because field strength does not depend on distance

A polar orbit with a 12-hour period that passes over the equator twice on each circuit

Any orbit at a height of exactly 1000 km, regardless of its direction or its period

An orbit that keeps the satellite directly above the poles as the Earth turns beneath it

An equatorial orbit with a 24-hour period in the same direction as the Earth's rotation

sixteen times as large

twice as large, since luminosity is directly proportional to the surface temperature

four times as large, since luminosity is proportional to the square of the temperature

eight times as large, since luminosity is proportional to the cube of the temperature

Above and to the right (cool and luminous)

Below and to the left (hot and dim)

At the top left only

Exactly on the main sequence

$v = \frac{H_0}{d}$

$d = H_0 v^2$

$v = H_0 d$

$v = H_0 d^2$