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3 years ago

What is the speed of a proton whose kinetic energy is 3.4 kev ?

1 answer:

8 0

The kinetic energy of the proton is 3.4 kev
1 kev = 1.602 × 10^-16 joules
therefore 3.4 kev is equivalent to;
3.4 × (1.602 ×10^-16)= 5.4468 × 10^-16 J
Kinetic energy is calculated by the formula 1/2mv² where m is the mass and v is the velocity.
Therefore V = √((2 × ( 5.4468×10^-16))/ (1.67 ×10^-27))
= 8.077 × 10^5 m/s

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Explanation:

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where

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$A_2$ is the cross-sectional area of the 2nd section of the pipe

$v_1$ is the velocity of the 1st section of the pipe

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In this problem we have:

$v_1=0.15 m/s$ is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

$d_2=0.74d_1$

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Now we can use Bernoulli's equation to find the pressure drop:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$

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$\rho=1025 kg/m^3$ is the blood density

$p_1,p_2$ are the initial and final pressure

So the pressure drop is:

$p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa$

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