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4 years ago

With a brief description, What are the 14 principles of management by fayol.

Engineering

1 answer:

Rudik [331]4 years ago

7 0

Answer:

is all about focus and unity.

Explanation:

All employees deliver the same activities that can be linked to the same objectives.

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A force of 250 N is acting on an area of 2 m2. The pressure is therefore:

lianna [129]

Answer:

125 Pascal

Explanation:

We know that

$Pressure=\frac{Force}{Area}$

Applying given values we get

$Pressure=\frac{250N}{2m^{2}}$

$Pressure=125\frac{N}{m^{2}}$

Pressure = 125 Pascal

7 0

3 years ago

The profile height of the tire sidewall (from bead to tread) is 117 mm.

Anna71 [15]

Answer:

498 mm

Explanation:

The height of the tire sidewall is ; 117 mm

The height of the rim of the tire is: 381 mm

Total height of the rim and tire together is : 117 + 381 = 498 mm

5 0

3 years ago

A particle moving in the x-y plane has a position vector given by r = 1.89t2i + 1.17t3j, where r is in inches and t is in second

ehidna [41]

Answer:

r=89.970 m

Explanation:

We need to calculate radius of curvature, using first and second derivatives of the functions x(t) and y(t).

From the given equation, we can get, that:

x(t)=1.89t^2 and y(t)=1.17t^3

Then, the first derivative:

x'(t)=2*1.89*t=3.78t and y'(t)=3*1.17*t^2=3.51t^2

The second derivatives are:

x''(t)=3.78 and y''(t)=3.51*2*t=7.02t

Equation for the radius of curvature, can be found as:

r=(x'^2+y'^2)^(3/2)/(x''y'-x'y'')

Note, that the denominator should be taken by its absolute value.

For the given time, we should calculate numerical values for the derivatives. For t=2.1 s:

x'(2.1)=7.938 m/s; x''(2.1)=3.78 m/s^2; y'(2.1)=15.4791 m/s and y''(2.1)=14.742 m/s^2

Using equation of the curvature radius and the values, we can get:

r=89.970 m

6 0

3 years ago

Read 2 more answers

A quantity of nitrogen gas in a piston–cylinder assembly undergoes a process at a constant pressureof 80 bar from 220 to 300 K.

notka56 [123]

Answer:

work done = 665.12 kJ/k.mol of nitrogen

dQ = 685.905 kJ/k.mol of nitrogen

Explanation:

given data

pressure = 80 bar

temperature t1 = 220 K

temperature t2 = 300 K

solution

first we get here work done as considering ideal gas condition

work done = P (v2-v1 ) = nR (t2-t1)

put here value

work done = 1 × 8.314 ×( 300 - 220 )

work done = 665.12 kJ/k.mol of nitrogen

and

now we get heat transfer by 1st law of thermodynamics that is

heat transfer dQ = dv + dw

dQ = Cv dT + dw

put here value and we get

dQ = $\frac{R}{\gamma -1 }$ × (t2-t1) + 665.12

dQ = $\frac{8.314}{1.4-1}$ × (300-220) + 665.12

dQ = 685.905 kJ/k.mol of nitrogen

8 0

3 years ago

The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface

slamgirl [31]

Answer:

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

F=ρ g A X

ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2 ⇒X=4 ft

A=8 x 10=80 $ft^2$

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

$F = 641,771.52 \dfrac{lb-ft}{s^2}$

4 0

4 years ago

With a brief description, What are the 14 principles of management by fayol.​

With a brief description, What are the 14 principles of management by fayol.