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3 years ago

Before you disconnect the service battery from the discharged battery, it is good practice to place a load across the

Engineering

2 answers:

Lilit [14]3 years ago

6 0

Answer:

it is true i just did this test

Explanation:

alexgriva [62]3 years ago

3 0

The answer should be true

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Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d

alexdok [17]

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number ( $Re$ ) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

$Re=\frac{\rho v D}{\mu}$

where $\rho$ is the density of the fluid, $\mu$ is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

$v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s$

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

$D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm$

8 0

3 years ago

A design team is working on creating a new locker organizer. They have

astraxan [27]

A. Present a Solution

8 0

3 years ago

Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t

Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

$h=\frac{2\sigma cos(\alpha )}{\rho gd}$

where

$\sigma$ is the surface tension of the liquid

$\alpha$ is the contact angle of the liquid

$\rho$ is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

$\sigma =7.28\times 10^{-2}N/m$

$\alpha =0$

$\rho _{w}=1000kg/m^3$

Applying the values we get

$h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm$

Part b) When the liquid is mercury:

For mercury and glass we have

$\sigma =485.5\times 10^{-3}N/m$

$\alpha =138^o$

$\rho _{w}=13.6\times 10^{3}kg/m^3$

Applying the values we get

$h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm$

The negative sign indicates that there is depression in mercury in the tube.

4 0

3 years ago

Use Routh's stability criterion to determine how many roots with positive real parts the following equations have:

Pavlova-9 [17]

Answer:

a) no roots not in LHP

b) 2 roots not in LHP

c) 2 roots not in the LHP

d) 2 roots not in the LHP

e) 2 roots not in LHP

Explanation:

$a) s^4 + 8s^3 + 32s^2 + 80s + 100 = 0\\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:32\:\:\:\:\:\:100\\s^3:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\\s^2:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:100\\s^1:\:\:\:80-\frac{800}{22} =43.6\\s^0:\:\:\:100$

No roots not in the LHP

$b) s^5 + 10s^4 + 30s^3 + 80s^2+344s + 480 =0 \\\\s^5:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:30\:\:\:\:\:\:344\\s^4:\:\:\:10\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\:\:\:\:\:\:480\\s^3:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:296\\s^2:\:\:\:-545\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:480\\s^1:\:\:\:490\\s^0:\:\:\:480$

2 roots not in the LHP

$c) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:7\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^3:\:\:\:2\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-2\\s^2:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^1:\:\:\:-4\\s^0:\:\:\:8$

There are roots in the RHP (not all coefficients are greater than 0).

2 roots not in the LHP

$d) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^3:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:20\\s^2:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:78\\s^1:\:\:\:-58\\s^0:\:\:\:78$

There are two sign changes in the first column of the Routh array.

2 roots not in the LHP

$e) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:6\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^3:\:\:\:4\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:12\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: new \:\:row \\s^2:\:\:\:3\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^1:\:\:\:12-\frac{100}{3}=-21.3 \\s^0:\:\:\:25$

2 roots not in LHP

check:

$a (s) = 0$ ⇒

$s^2 = -3 \limits^+_- 4j = 5e^{j(\pi \limits^+_- 0.92)}\\\\s = \sqrt5 e^{j( \frac{\pi}{2} \limits^+_- 0.46)+n\pi j},\:\:\:\:\: n= 0, 1\\$

3 0

3 years ago

When working cattle through a chute, stop cattle by moving…point of balance.

timofeeve [1]

When the handler has reached the point of balance of the animal, they should stop walking so that they may move only one animal at a time. The point of balance for cattle that are being worked with in limited places such as races and chutes is often found at the animal's shoulder, as seen by the diagrams.

This is further explained below.

<h3>What is a chute?</h3>

Generally, Chutes are channels, planes, or passageways that are either vertical or slanted and through which things may be transported using only gravity.

In conclusion, When the handler has reached the point where the animal's point of balance has been crossed, they should stop walking so that they may move just one animal.

As seen in the illustrations, the point of balance for cattle being worked with in restricted spaces like races and chutes is often located at the animal's shoulder.