Answer:
The magnitude of force the fluid leaves the needle is 3.75 x 10⁻³ N
Explanation:
Given;
applied force to the plunger, F = 6 N
radius of the plunger, r = 0.40 cm = 4 x 10⁻³ m
radius of the needle, r = 0.010 cm = 1 x 10⁻⁴ m
Pressure of the fluid is given as;
P = F / A
where;
A is area, = πr²
[tex]P = \frac{F_{plunger}}{\pi r_{plunger}^2} = \frac{F_{needle}}{\pi r_{needle}^2}[/tex]
if pressure of the fluid is constant, the force applied to the plunger and force the fluid leaves the needle is given as;
[tex]F_{needle} = \frac{F_{plunger}*r_{needle}^2}{r_{plunger}^2} \\\\F_{needle} = \frac{6 *(1*10^{-4})^2 }{(4*10^{-3})^2} = 3.75*10^{-3}\ N[/tex]
Therefore, the magnitude of force the fluid leaves the needle is 3.75 x 10⁻³ N
