Read the following passage and answer the accompanying questions - Leaving Cert Physics - Question 11 - 2019

(i) Transformer (ii) Rectifier/Diode

Hooke's Law states that the restoring force is proportional to the displacement from equilibrium:

$F = -kx$

where $F$ is the restoring force, $k$ is the spring constant, and $x$ is the displacement.

First, calculate the change in momentum:

$ext{Change in momentum} = m(v_f - v_i)$

where:

• $m = 0.110$ kg (mass in kg),
• $v_f = -4$ m s⁻¹ (final velocity, negative as it's in the opposite direction),
• $v_i = 4$ m s⁻¹ (initial velocity).

Calculating gives:

$ext{Change in momentum} = 0.110 kg imes (-4 - 4) = 0.110 kg imes -8 = -0.88 kg ext{ m s}^{-1}$

Now, using the impulse-momentum theorem:

$F imes t = ext{Change in momentum}$

where $t = 0.2 s$.

Solving for $F$ gives:

F = rac{ ext{Change in momentum}}{t} = rac{-0.88 kg ext{ m s}^{-1}}{0.2 s} = -4.4 N

Thus, the force exerted by the wall is 4.4 N in the opposite direction.

In the ray diagram, depict the lens and the object being magnified. Draw the following:

1. Object placed within the focal length of the lens.
2. Two rays emanating from the top of the object: one passing through the optical center of the lens, and the other refracted downward towards the focal point.
3. The point where the extended rays appear to diverge from creates the image.

Indicate the 'upright' nature of the formed image on the diagram.

The nuclear equation for the fission reaction is:

Using the mass defect, we can calculate the energy released as follows:

• Loss in mass = 3.3682 x $10^{-28}$ kg
• Using Einstein's equation:

$E = mc^2$

First, calculate:

$E = 3.3682 imes 10^{-28} kg imes (2.9979 imes 10^8 m/s)^2$

This results in: $E = 3.0271 imes 10^{-11} J$

The energy is released in the form of kinetic energy during the fission process.