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3 years ago

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7. Plasma from blood (density = 1025 kg/m3) flows along a vertical channel in a steady, incompressible, fully developed laminar

film of thickness h. (i) Simplify the continuity and Navier-Stokes equations to model this flow field. (ii) Obtain expressions for the velocity profile, (iii) the shear stress distribution, (iv) the volume flow rate, and (v) the average velocity. (vi) Relate the plasma film thickness to the volume flow rate per unit depth of surface normal to the flow. (vii) Calculate the volume flow rate for h = 0.5 mm, flowing down a surface b = 2 cm wide.

1 answer:

Goryan [66]3 years ago

6 0

Answer: Q = 8.37×10^-9L/s

Explanation: since the depth of the surface is normal to the flow, ø = 90

Find the attached file for the solution

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Three Carnot engines operate between the following temperature limits.

Kipish [7]

Answer:c>a>b

Explanation:

Given

All the engine extracts same amount of heat(Q) from High-temperature reservoir

For a) 400 and 500 K

$\eta _{engine}=1-\frac{T_L}{T_H}$

$\eta _{engine}=\frac{work\ supplied}{heat\ supplied}$

$1-\frac{400}{500}=\frac{W_a}{Q}$

$W_a=\frac{Q}{5}$

For b)500 K and 600K

$1-\frac{500}{600}=\frac{W_b}{Q}$

$W-b=\frac{Q}{6}$

For c) 400 K and 600 K

$1-\frac{400}{600}=\frac{W_c}{Q}$

$W_c=\frac{2Q}{3}$

So c will give the highest amount of work

c>a>b

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3 years ago

The cube has a mass of 72.9 g. What substance is this cube composed of?

aalyn [17]

I Need More Details.
But here is somthing i found maybe the same.
brainly.com/question/1354966

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3 years ago

A long, thin solenoid has 390 turns per meter and a radius of 1.20 cm. The current in the solenoid is increasing at a uniform ra

gtnhenbr [62]

To solve this problem it is necessary to apply the concepts related to Faraday's law and the induced emf.

By definition the induced electromotive force is defined as

$\int E dl = -\frac{d\phi}{dt}$

$\int E dl = -(\frac{dB}{dt})A$

Where,

$\phi =$ Electric field

B = Magnetic Field

A = Area

At the theory the magnetic field is defined as,

$B = \mu_0 NI$

Where,

N = Number of loops

I = current

$\mu_0 =$ Permeability constant

We know also that the cross sectional area, is the area from a circle, and the length is equal to the perimeter then

A = \pi r^2

l = 2\pi r

Replacing at the previous equation we have that

$E (2\pi r) = \mu_0 n (\frac{di}{dt})(\pi R^2)$

Where,

R = Radius of the solenoid

r = The distance from the axis

Re-arrange to find the current in function of time,

$\frac{di}{dt} = \frac{Er}{\mu_0 NR^2}$

Replacing our values we have

$\frac{di}{dt} = \frac{(8.00*10^{-6})(0.0348)}{(4\pi*10^{-7})(390)(1.2*10^-2)^2}$

$\frac{di}{dt} = 3.94487A/s$

8 0

3 years ago

If the direction of the position is north and the direction of the velocity is up, then what is the direction of the angular mom

Kay [80]

Answer:

the direction of angular momentum = EAST

Explanation:

given

Direction of position = r = north

Direction of velocity = v = up

angular momentum = L = m(r x v)

where m is the mass, r is the radius, v is the velocity

utilizing the right hand rule, the right finger heading towards the course of position vector and curl them toward direction of velocity, at that point stretch thumb will show the bearing of the angular momentum.

then L = north x up = East

6 0

3 years ago

Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with

IgorLugansk [536]

(a) $2.5\cdot 10^{-6}eV$

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

For the microwave photon,

$\lambda=50.00 cm = 0.50 m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV$

(b) $2.5 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the visible light photon,

$\lambda=500 nm = 5 \cdot 10^{-7}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV$

(c) $2500 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the x-ray photon,

$\lambda=0.5 nm = 5 \cdot 10^{-10}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV$

6 0

3 years ago