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3 years ago

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Consider blood (density 1025 kg/m3) flowing through an artery with circular cross section that looks like this: Ignore viscosity

. Assume that the change in height is negligible between regions A, B, and C.13. What is the mass flow rate through region A

1 answer:

Klio2033 [76]3 years ago

8 0

The given question is incomplete. The complete question is attached with an image below.

Explanation:

Mass flow rate through region A will be calculated as follows.

Rate = $\rho \times A \times \mu$

= $\rho \times \pi \times R^{2} \times \mu$

= $1025 \times 3.14 \times (0.5 \times 10^{-2})^{2} \times 30 \times 10^{-2}$

= $24138.75 \times 10^{6} kg/s$

Therefore, we can conclude that the mass flow rate through region A is $24138.75 \times 10^{6} kg/s$ .

You might be interested in

(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rat

vitfil [10]

Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i) the gradient of f = ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is ñ∙∇F = 4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+ ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+ (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have

∇f(x, y) = -1·i -3·j that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z) = ∇F(x, y, z) = = ∂f/∂x i+ ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore

ñ = ⅟4(2·i +3·j -√3·k)

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

8 0

3 years ago

A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2

Pepsi [2]

Work needed = 23,520 J

<h3> Further explanation </h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object

Work is the product of force with the displacement of objects.

Can be formulated

W = F x d

W = Work, J, Nm

F = Force, N

d = distance, m

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

8 0

2 years ago

when energy is transferred from one part of a system to another, some of the energy is lost during the transfer and cannot be us

AlladinOne [14]

Sentences A and D describe examples of energy transformation.Heat is produced when a car's tires rub against the pavement and as electricity passes across power wires, they become hotter.

<h3>What is the law of conservation of energy?</h3>

According to the law of conservation of energy, the energy of an isolated system stays unchanged throughout time.it is said to be conserved.

Energy cannot be created nor destroyed and can be transferred from one form to the other form.

The complete question is

"When energy is moved from one component of a system to another, some of the energy is lost and cannot be used as planned.

Which two statements provide examples?

A. Friction between a car's tires and the road produces heat.

B. Sunlight strikes a solar panel, generating electricity.

c. Stereo speakers emit a sound when powered by electrical energy,

D. Wind moves a turbine, generating electricity.

I E. Power lines heat up as electricity flows through them."

Some of the energy wasted during the movement of energy from one section of a system to another is heat is produced by friction between a car's tires and the road and as electricity passes via power lines, they heat up.

Hence, sentences A and D describe examples of energy transformation.

To learn more about the law of conservation of energy refer to the link;

brainly.com/question/2137260

#SPJ1

5 0

1 year ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

a man crossed a road 8.25m wide at a speed of 2.01m/s,how long does it take to get man to cross the road

Svetradugi [14.3K]

Answer:

t = 4.1 seconds

Explanation:

It is given that,

Width of road which is to be crossed by a man is 8.25 m, it means it is distance to be covered.

Speed of man is 2.01 m/s

We need to find the time taken by the man to cross the road. It is a concept of speed. Speed of a person is given by total distance covered divided by time taken. So,

$v=\dfrac{d}{t}$

t is time taken

$t=\dfrac{d}{v}\\\\t=\dfrac{8.25}{2.01}\\\\t=4.1\ s$

So, the time taken by the man to cross the road is 4.1 seconds.

5 0

3 years ago