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A wide hose pipe is held horizontally by a fireman. It delivers water through a nozle at one lirte per second. On increasing the pressure, this increass to two litres per second. The fireman has now to
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push forward four times as hard
Rate of change of mass, $\frac{d m}{d t}=a v \rho$
$\therefore F=\frac{v d m}{d t}=(a v \rho) v=a v^2 \rho$
Volume of liquid flowing per second $=a v$
This is proportional to velocity.
By doubling the volume of liquid flow per second, the force becomes four times. When liquid flows forward, the hose pipe tends to come backward. So, to keep it intact, it should be pushed forward. Thus, the hose pipe should be pushed forward four times.
$\therefore F=\frac{v d m}{d t}=(a v \rho) v=a v^2 \rho$
Volume of liquid flowing per second $=a v$
This is proportional to velocity.
By doubling the volume of liquid flow per second, the force becomes four times. When liquid flows forward, the hose pipe tends to come backward. So, to keep it intact, it should be pushed forward. Thus, the hose pipe should be pushed forward four times.
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