Answer:
13.65 m/s
Explanation:
Taken the speed of the liquid in the syringe to zero and using Bernoulli's principle
ΔP ( change in pressure) = Force applied / area = 2.05 N / 2.20 × 10 ⁻⁵ m² = 93181.82 N / m²
ΔP = 1/2 ρv² where v is the speed of medicine at the mouth of the syringe and rho is density of water which 1000 kg/m³
93181.82 N / m² = 0.5 × 1000 × v²
v² = (93181.82 N / m²) / (0.5 × 1000 ) = 13.65 m/s
The speed of the medicine as it leaves the needle's tip is 13.64 m/s.
Given data:
The cross-sectional area of the barrel is, [tex]A = 2.20 \times 10^{-5} \;\rm m^{2}[/tex].
The cross-sectional area of needle is, [tex]a = 1.00 \times 10^{-8} \;\rm m^{2}[/tex].
The weight of arrow is, W = 2.05 N.
Let us consider the liquid in the syringe to be at rest. Then the speed of liquid in the syringe will be zero. Then applying the Bernoulli's concept for the pressure change as,
[tex]\Delta P = \dfrac{1}{2} \times \rho \times v^{2}[/tex] ...........................................(1)
Here,
v is the speed of medicine at the mouth of the syringe.
[tex]\rho[/tex] is density of water which 1000 kg/m³.
And change in pressure is also expressed as,
[tex]\Delta P = \dfrac{W}{A}[/tex]
Solving as,
[tex]\Delta P = \dfrac{2.05}{2.20 \times 10^{-5}}\\\\\Delta P = 9.31 \times 10^{5} \;\rm Pa[/tex]
Substitute the values in equation (1) as,
[tex]9.31 \times 10^{4} = \dfrac{1}{2} \times 1000 \times v^{2}\\\\v=\sqrt{\dfrac{2 \times 9.31 \times 10^{4}}{1000}}\\\\v=13.64 \;\rm m/s[/tex]
Thus, the speed of the medicine as it leaves the needle's tip is 13.64 m/s.
Learn more about the Bernoulli's concept here:
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