Fill in the blanks:
$(a)$ $A$ load of $3 \, kg$ is suspended from a rope of $6 \, kg$. The tension at the top end of the rope is ...........
$(b)$ The effect produced by a force $F$ acting for a time $\Delta t$ is the same as that produced by a force $2F$ acting for a time ...........
$(c)$ Friction ........... due to excessive ironing.

(N/A) The total mass supported at the top end is the sum of the load mass and the rope mass: $M = 3 \, kg + 6 \, kg = 9 \, kg$. The tension $T$ at the top is $T = Mg = 9 \times 9.8 = 88.2 \, N$ (or simply $9g \, N$).
$(b)$ The impulse-momentum theorem states that the change in momentum is equal to the impulse,$J = F \Delta t$. For the same effect (same impulse),$F_1 \Delta t_1 = F_2 \Delta t_2$. Given $F_1 = F$,$\Delta t_1 = \Delta t$,and $F_2 = 2F$,we have $F \Delta t = 2F \times \Delta t_2$,which gives $\Delta t_2 = \frac{\Delta t}{2}$.
$(c)$ Excessive ironing increases the surface roughness or creates microscopic welds between surfaces,which increases friction.