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zlopas [31]
3 years ago
12

lood flows through a section of a horizontal artery that is partially blocked by a deposit along the artery wall. As a hemoglobi

n molecule moves from the narrow region into the wider region, its speed changes from v2 = 0.800 m/s to v1 = 0.475 m/s. What is the change in pressure, P1 - P2, that it experiences? The density of blood is 1060 kg/m3.
Physics
1 answer:
uysha [10]3 years ago
5 0

Answer:

P_1 - P_2 =219.62\ Pa

Explanation:

given,

density of blood = 1060 kg/m³

v₂ = 0.800 m/s

v₁ = 0.475 m/s

change in pressure calculation

using Bernoulli's equation

P_1 +\dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2

both are at same level

P_1 - P_2 =\dfrac{1}{2}\rho (v_2^2- v_1^2)

P_1 - P_2 =\dfrac{1}{2}1060\times (0.8^2- 0.475^2)

P_1 - P_2 =219.62\ Pa

the change in pressure is equal to 219.62 Pa

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A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
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(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

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