An electromagnetic relay exerts a force of 12N and move a satt tron armature furough a distance of 1.5 cm in 0.5 s. Find the power Consumed.

The work done ($W$) is the product of force ($F$) and distance ($d$).

$$W = F \times d$$

Given $F = 12\, \text{N}$ and $d = 1.5\, \text{cm} = 0.015\, \text{m}$.

$$W = 12\, \text{N} \times 0.015\, \text{m}$$ $$W = 0.18\, \text{J}$$

Power ($P$) is the work done ($W$) divided by the time ($t$).

$$P = \frac{W}{t}$$

Given $W = 0.18\, \text{J}$ and $t = 0.5\, \text{s}$.

$$P = \frac{0.18\, \text{J}}{0.5\, \text{s}}$$ $$P = 0.36\, \text{W}$$

The power consumed is $\boxed{0.36\, \text{W}}$.

2. Accelerating Force of a Car

Given mass ($m$) = $1.7\, \text{Mg}$.
Since $1\, \text{Mg} = 1000\, \text{kg}$:

$$m = 1.7 \times 1000\, \text{kg}$$ $$m = 1700\, \text{kg}$$

The accelerating force ($F$) is the product of mass ($m$) and acceleration ($a$).

$$F = m \times a$$

Given $m = 1700\, \text{kg}$ and $a = 3\, \text{m/s}^2$.

$$F = 1700\, \text{kg} \times 3\, \text{m/s}^2$$ $$F = 5100\, \text{N}$$

The accelerating force is $\boxed{5100\, \text{N}}$.

3. Battery Terminal p.d. and Load Resistance

The terminal p.d. ($V$) is given by the EMF ($\mathcal{E}$) minus the voltage drop across the internal resistance ($Ir$).

$$V = \mathcal{E} - Ir$$

Given $\mathcal{E} = 12\, \text{V}$, $I = 10\, \text{A}$, and $r = 0.2\, \Omega$.

$$V = 12\, \text{V} - (10\, \text{A} \times 0.2\, \Omega)$$ $$V = 12\, \text{V} - 2\, \text{V}$$ $$V = 10\, \text{V}$$

The p.d. at the battery terminal is $\boxed{10\, \text{V}}$.

Using Ohm's Law for the external circuit, the terminal p.d. ($V$) is equal to the current ($I$) multiplied by the load resistance ($R$).

$$V = IR$$

We found $V = 10\, \text{V}$ and given $I = 10\, \text{A}$.

$$10\, \text{V} = 10\, \text{A} \times R$$ $$R = \frac{10\, \text{V}}{10\, \text{A}}$$ $$R = 1\, \Omega$$

The resistance of the load is $\boxed{1\, \Omega}$.