Please solve those questions in English

The gradient $m$ is given by the formula $m = \frac{\Delta V}{\Delta I}$. Using the points $(0.0 \text{ A}, 0.0 \text{ V})$ and $(7.1 \text{ A}, 10.0 \text{ V})$ from the table (representing the best straight line passing through the origin):

$$m = \frac{10.0 \text{ V} - 0.0 \text{ V}}{7.1 \text{ A} - 0.0 \text{ A}} = \frac{10.0}{7.1} \approx 1.408 \text{ V/A}$$

The gradient of a potential difference ($V$) versus current ($I$) graph represents the resistance of the conductor.
The gradient is approximately $\boxed{\text{1.41 } \Omega}$. Its significance is the resistance of the conductor.

8. (a) (iv)

From Ohm's Law, $V = IR$. We calculated the resistance $R \approx 1.408 \Omega$ from the gradient. We want to find the current $I$ when $V = 7.5 \text{ V}$.

$$I = \frac{V}{R}$$

$$I = \frac{7.5 \text{ V}}{1.408 \Omega} \approx 5.3267 \text{ A}$$

The current in the circuit when the potential difference is 7.5 V is approximately $\boxed{\text{5.33 A}}$.

8. (b) (i)
Electromotive force (e.m.f.) is the energy supplied by a source (such as a battery) per unit charge to drive the charge around a complete circuit. It is measured in volts (V) and represents the potential difference across the terminals of the source when no current is flowing.

8. (b) (ii)
The resistance of a wire is given by the formula $R = \rho \frac{L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area. If the wire is replaced by a thinner wire of the same material and length, its cross-sectional area $A$ decreases. A decrease in $A$ leads to an increase in resistance $R$. According to Ohm's Law ($I = \frac{V}{R}$), if the resistance $R$ increases and the battery voltage $V$ remains constant, the current $I$ will decrease.

8. (c) (i)
The function of a fuse is to protect electrical circuits from overcurrents. It melts and breaks the circuit if the current exceeds a safe level, preventing damage to appliances and reducing fire hazards. It is connected in series with the live wire of the circuit.

8. (c) (ii)
A fuse contains a thin wire made of a material with a low melting point. When the current flowing through the circuit exceeds the fuse's rated value, the wire heats up due to the Joule heating effect ($P = I^2 R$). This heat causes the fuse wire to melt, creating an open circuit. This interruption stops the flow of current, thereby protecting appliances and preventing electrical hazards.