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suter [353]
4 years ago
7

if change in blood pressure between the brain and the feet is 1.88×10^4 pa .what will be the volume flow rate from head to feet

of yhe 6feet person if his radius is 23cm​
Physics
1 answer:
klio [65]4 years ago
4 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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An object is thrown upward from the top of a 128​-foot building with an initial velocity of 112 feet per second. The height h of
lorasvet [3.4K]

Answer:

Time, t = 8 seconds

Explanation:

An object is thrown upward from the top of a 128​-foot building with an initial velocity of 112 feet per second. The height h as a function of time t is given by :

h=-16t^2+112t+128

We need to find the time when the object will hit the ground. When it will hit the ground, h = 0

So,

-16t^2+112t+128=0

On solving the above quadratic equation, we get the value of t = 8 seconds. So, after 8 seconds the object will hit the ground. Hence, this is the required solution.

6 0
4 years ago
One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

F_y = 945N

In such a way that the material is more resistant to tensile strength than shear force.

6 0
3 years ago
On a bet, you try to remove water from a glass by blowing across the top of a vertical straw immersed in the water. What is the
MArishka [77]

Answer:

       v₂ = 0.56 m / s

Explanation:

This exercise can be done using Bernoulli's equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Where points 1 and 2 are on the surface of the glass and the top of the straw

The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of ​​the straw the velocity of the surface of the vessel is almost zero v₁ = 0

The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m

We substitute in the equation

         P_{atm} + ρ g y₁ = P_{atm} + ½ ρ v₂² + ρ g y₂

         ½ v₂² = g (y₂-y₁)

        v₂ = √ 2 g (y₂-y₁)

Let's calculate

        v₂ = √ (2 9.8 1.6 10⁻²)

       v₂ = 0.56 m / s

5 0
3 years ago
Explain the process of refraction using the key words: Speed, normal, angle
Tanzania [10]
Https://www.physicsclassroom.com/class/refrn/Lesson-2/The-Angle-of-Refraction
6 0
3 years ago
Repeating anyone ineed answers plsss?
hichkok12 [17]

Answer:

the answer of question no5 is b

3 0
3 years ago
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