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3 years ago

10

A bona fide established commercial marketing agency is a business which is specifically devoted to public relations, advertising

and promoting the services of a client. True or False

1 answer:

Katarina [22]3 years ago

5 0

Answer:

True

Explanation:

Bona Fide is a Latin term which means in good faith or without any intention to deceive. The business established on a bona fide basis means that there is an absence of fraud. The marketing agency has devoted its services to public relations, advertising and promoting the services of clients. There is no intention of fraud in the business.

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A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the

Margaret [11]

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

$C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2)$

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

$T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

$C_p = 520 \:\: J/kg.K$

So, the inlet temperature of argon is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$

$T_{in} = \frac{1}{2} \times 119 + 300$

$T_{in} = 360K$

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

$C_p = 5193 \:\: J/kg.K$

So, the inlet temperature of helium is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$

$T_{in} = \frac{1}{2} \times 12 + 300$

$T_{in} = 306K$

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

$C_p = 1039 \:\: J/kg.K$

So, the inlet temperature of nitrogen is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$

$T_{in} = \frac{1}{2} \times 60 + 300$

$T_{in} = 330K$

Note: Answers are rounded to the nearest whole numbers.

5 0

2 years ago

During the recovery of a cold-worked material, which of the following statement(s) is (are) true?

Len [333]

Answer:

Some of the internal strain energy is relieved.

There is some reduction in the number of dislocations.

The electrical conductivity is recovered to its precold-worked state.

The thermal conductivity is recovered to its precold-worked state

Explanation:

The process of the recovery of a cold-worked material happens at a very low temperature, this process involves the movement and annihilation of points where there are defects, also there is the annihilation and change in position of dislocation points which leads to forming of the subgrains and the subgrains boundaries such as tilt, twist low angle boundaries.

4 0

3 years ago

Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.

Novosadov [1.4K]

Answer:

i) SF: $v(x) = \frac{(w_0* x )^2}{2L}$

ii) BM : $= \frac{(w_0*x)^3}{6L}$

Explanation:

Let's take,

$\frac{y}{w_0} = \frac{x}{L}$

Making y the subject of formula, we have :

$y = \frac{x}{L} * w_0$

For shear force (SF), we have:

This is the area of the diagram.

$v(x) = \frac{1}{2} * y = \frac{1}{2} * \frac{x}{L} * w_0$

$= \frac{(w_0* x )^2}{2L}$

The shear force equation =

$v(x) = \frac{(w_0* x )^2}{2L}$

For bending moment (BM):

$BM = v(x) * \frac{x}{3}$

$= \frac{(w_0* x )^2}{2L} * \frac{x}{3}$

$= \frac{(w_0*x)^3}{6L}$

The bending moment equation =

$= \frac{(w_0*x)^3}{6L}$

5 0

3 years ago

a vertical cylindrical container is being cooled in ambient air at 25 °C with no air circulation. if the initial temperature of

Sloan [31]

Answer:

the surface heat-transfer coefficient due to natural convection during the initial cooling period. = 4.93 w/m²k

Explanation:

check attachement for answer explanation

7 0

2 years ago

Read 2 more answers

A heat engine operates between 2 reservoirs at TH and 18oC. The heat engine receives 17,000 kJ/h from the high temperature reser

lisabon 2012 [21]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

4 0

2 years ago