Answer
$a.\quad 0.25\mathrm{N}$
$ b.\quad$more than 2.5 N
$c.\quad 2.8\mathrm{N}$
Work Step by Step
If fuel is expelled with the speed $v$ and at the rate $\Delta m/\Delta t$.
The thrust experienced is:
thrust $=(\displaystyle \frac{\Delta m}{\Delta t})v \qquad$9-19
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$a.$
Given the mass/meter and meter/second rates,
$\displaystyle \frac{\Delta m}{\Delta t}$=$\displaystyle \frac{\Delta m}{\Delta x}\cdot\frac{\Delta x}{\Delta t}=(0.13 \mathrm{k}\mathrm{g}/\mathrm{m} )(1.4 \mathrm{m}/\mathrm{s})$
So,
thrust $=(\displaystyle \frac{\Delta m}{\Delta t})v=(0.13 \mathrm{k}\mathrm{g}/\mathrm{m} )(1.4 \mathrm{m}/\mathrm{s})^{2}=0.25\mathrm{N}$
$b.$
The scale reads the weight of the rope on the scale plus the thrust due to the falling rope. It reads more than 2.5 N.
$c.$
Scale reading $=$ thrust + weight
$=0.25\mathrm{N}+2.5\mathrm{N}=2.8\mathrm{N}$
