Ask question

Ask question

All categories

Vladimir79 [104]

2 years ago

6

Which is the most effective way to reduce discrimination at the workplace? A. Appoint employees of a single race or religion. B.

Create a group according to the background of the employees. C. Conduct awareness training programs for all employees. D. Ask employees to choose their team members

2 answers:

sveticcg [70]2 years ago

6 0

Answer:

C is the correct answer coz the others don't make sense

Harlamova29_29 [7]2 years ago

5 0

The correct answer to your question is c

You might be interested in

A scale model is 4th the size of the pump. Determine the power ratio of the pump and its scale model if the ratio of the heads i

vredina [299]

Given:

size of scale model = 4(size of pump)

power ratio of pump and scale model = 5:1

Solution:

Let the diameter of scale model and pump be $d_{s}$ and $d_{p}$ respectively

and head be $H_{s}$ and $H_{p}$ respectively

Now, power, P is given as a function of head(H) and dischagre(Q)

P = $\rho gQH$ (1)

From eqn (1):

$P \propto QH$

and

$QH \propto \sqrt{H}D^{2}$

So,

$P \propto H^{\frac{3}{2}} D^{2}$

Therefore,

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{1^{2}\times 5^{\frac{3}{2}}}{4^{2}\times 1^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{5\sqrt{5}}{16}$

${P_{s}}:{P_{p}}$ = ${5\sqrt{5}}:{16}$

8 0

3 years ago

How can you use the IPDE Process to help protect motorcyclists while driving?

gladu [14]

Answer: you can use the IPDE process to protect motorcyclists while driving by identifying when a motorcycle is present, predicting what it will do next, deciding how to react to the other drivers decision, and exacting that decision.

I hope this helped

3 0

3 years ago

Why the velocity potential Φ(x,y,z,t) exists only for irrotational flow

Black_prince [1.1K]

Answer:

$\omega_y,\omega_x,\omega_Z$ all are zero.

Explanation:

We know that if flow is possible then it will satisfy the below equation

$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0$

Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.

And velocity potential function $\phi$ given as follows

$u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}$

Rotationality of fluid is given by $\omega$

$\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z$

$\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x$

$\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y$

So now putting value in the above equations ,we will find

$\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},$

$\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}$

So $\omega_y$ =0

Like this all $\omega_y,\omega_x,\omega_Z$ all are zero.

That is why velocity potential flow is irroational flow.

5 0

4 years ago

Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street r

irakobra [83]

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area = 1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000 × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

4 0

3 years ago

An automobile tire with a volume of 0.8 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if t

saveliy_v [14]

Answer:

2.83 kg

Explanation:

Given:

Volume, V = 0.8 m³

gage pressure, P = 200 kPa

Absolute pressure = gage pressure + Atmospheric pressure

= 200 + 101 = 301 kPa = 301 × 10³ N/m²

Temperature, T = 23° C = 23 + 273 = 296 K

Now,

From the ideal gas equation

PV = mRT

Where,

m is the mass

R is the ideal gas constant = 287 J/Kg K. (for air)

thus,

301 × 10³ × 0.8 = m × 287 × 296

or

m = 2.83 kg

3 0

4 years ago