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ololo11 [35]
3 years ago
6

ΔP = 1.88 x 10^4 Pa. Use this answer to estimate the volume flow rate of blood from the head to the feet of a six-foot-tall pers

on assuming an effective radius of 23cm
Physics
1 answer:
Sveta_85 [38]3 years ago
7 0

Answer: 3765.66 \frac{m^{3}}{s}

Explanation:

We can solve this problem using the <u>Poiseuille equation</u>:

Q=\frac{\pi r^{4}\Delta P}{8\eta L}

Where:

Q  is the Volume flow rate

r=23 cm \frac{1 m}{100 cm}=0.23 m  is the effective radius

L=6 ft \frac{0.3048 m}{1 ft}=1.8288 m  is the length

\Delta P=1.88(10)^{4} Pa  is the difference in pressure

\eta=3(10)^{-3} Pa.s is the viscosity of blood

Solving:

Q=\frac{\pi (0.23 m)^{4}(1.88(10)^{4} Pa)}{8(3(10)^{-3} Pa.s)(1.8288 m)}

Q=3765.66 \frac{m^{3}}{s}

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A professional baseball player can throw the ball around 45 m/s if the distance between the pitcher and the batter is 18.39 m. H
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Answer:

The time taken for the ball to get to the batter is 0.41 s.

Explanation:

Given;

initial velocity of the baseball, u = 45 m/s

horizontal distance between the pitcher and the batter, X = 18.39 m

The horizontal distance or range of a projectile is given as;

X = ut

where;

t is the time of flight

u is the initial velocity

t = X / u

t = 18.39 / 45

t = 0.41 s

Therefore, the time taken for the ball to get to the batter is 0.41 s.

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Find the work done on a 50 Kg student by the elevator in 2 seconds, if the elevator is :
11Alexandr11 [23.1K]

The work done on a 50 Kg student by the elevator in 2 seconds if the elevator is accelerating upwards from rest at a rate of 2 ms⁻² would be

<h3>What is work done?</h3>

The total amount of energy transferred when a force is applied to move an object through some distance.

The work done is the multiplication of applied force with displacement.

Work Done = Force × Displacement

As given in the problem we have to find the work done  on a 50 Kg student by the elevator in 2 seconds if the elevator is  accelerating upwards from rest at a rate of 2 ms⁻²,

Lets us first calculate the displacement of the elevator in 2 seconds if it is accelerating upwards from rest at a rate of 2 ms⁻².

s = ut + 0.5at²

  = 0 + 0.5×2×2²

  = 4 meters

The work done on the elevator = mgh

                                                    =50×9.81×4

                                                    =1962 joules

Thus, the work done on the elevator would be 1962 joules.

To learn more about work done, refer to the link;

brainly.com/question/13662169

#SPJ1

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