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1 year ago

Where can you go to find the system requirements?

Engineering

1 answer:

SSSSS [86.1K]1 year ago

8 0

Answer:c

Explanation:

yea, jus go to the systems help center then youll be confronted with a talks. person n they'll walk u threw it.

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What are the units or dimensions of the shear rate dv/dy (English units)? Then, what are the dimensions of the shear stress τ= μ

swat32

Answer:

1) Dimensions of shear rate is $[T^{-1}]$ .

2)Dimensions of shear stress are $[ML^{-1}T^{-2}]$

Explanation:

Since the dimensions of velocity 'v' are $[LT^{-1}]$ and the dimensions of distance 'y' are $[L]$ , thus the dimensions of $\frac{dv}{dy}$ become

$\frac{[LT^{-1}]}{[L]}=[T^{-1}]$ and hence the units become $s^{-1}$ .

Now we know that the dimensions of coefficient of dynamic viscosity $\mu$ are $[ML^{-1}T^{-1}]$ thus the dimensions of shear stress can be obtained from the given formula as

$[\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}]$

Now we know that dimensions of momentum are $[MLT^{-1}]$

The dimensions of $Area\times time$ are $[L^{2}T]$

Thus the dimensions of $\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}]$

Which is same as that of shear stress. Hence proved.

4 0

3 years ago

Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−

sergeinik [125]

Answer:

Power output, $P_{out} = 178.56 kW$

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, $T = 350^{\circ}C$

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, $\dot{m} = 0.1 kg.s^{- 1}$

Diameter of exhaust pipe, $d_{h} = 15 cm = 0.15 m$

Pressure at exhaust, P' = 50 kPa

temperature, T' = $100^{\circ}C$

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

$\dot{m} = \frac{Av_{i}}{V_{1}}$

$0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}$

$0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}$

$v_{i} = 3.986 m/s$

Now, eqn for compressible fluid:

$\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}$

Now,

$\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}$

$\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}$

$\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}$

$v_{e} = 19.33 m/s$

Now, the power output can be calculated from the energy balance eqn:

$P_{out} = -\dot{m}W_{s}$

$P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}$

$P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW$

4 0

2 years ago

Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly fr

katrin [286]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

$\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}$

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

$\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m$

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

$n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times$

c) By using the equation of the ideal gases you obtain:

$PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules$

5 0

2 years ago

If Ella decided to become a children’s doctor, what career cluster would she belong in?

vodomira [7]

D . Health Services

8 0

2 years ago

In a surface grinding operation, the wheel diameter = 8.0 in, wheel width = 1.0 in, wheel speed = 6000 ft/min, work speed = 40 f

Katen [24]

Answer: the answer will be d because it is the right one to be

Explanation:

7 0

3 years ago