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Sophie [7]
3 years ago
13

Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t

he artery to 74 % of its diameter. What pressure drop occurs when the blood reaches this narrow section? Ignore the blood viscosity.
Physics
1 answer:
sp2606 [1]3 years ago
8 0

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

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

v_2 is the velocity of the 2nd section of the pipe

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

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

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

where

\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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Answer:

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Explanation:

given,

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The block is made of A) Tin, as its specific heat capacity is 0.225 J/(g^{\circ}C)

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by \Delta T, according to the following equation :

Q=mC_s \Delta T

where  C_s is the specific heat capacity of the material.

In this problem, we have:

m = 2 kg = 2000 g is the mass of the unknown material

Q = 450 J is the amount of energy supplied to the block

\Delta T = 1^{\circ}C is the change in temperature of the material

Solving the equation for C_s, we can find the specific heat capacity of the unknown sample:

C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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